I am a Masters year Theoretical Physics student at the University of Birmingham. I am passionate about all things physics and am particularly keen on anything with a strong mathematical element. I also have a strong interest in the communication of science and maths to people of a wide range of abilities. I'm always looking for new ways to think about scientific concepts to make them more easily accessible. I believe that science and maths don't need to be as exclusive as they currently are!

During my time in the Air Cadets, I designed and delivered a teaching programme for teenagers of varying ages and abilities in order to teach subjects related to the principles of flight, history, navigation amongst other subjects. Whilst studying at university, I was also awarded the Forresters UoB Clear Direction prize for best scientific communicator. As a tutor, I feel that I can bring these skills together to make even the toughest scientific and mathematical concepts accessible in the lesson space.

I am a Masters year Theoretical Physics student at the University of Birmingham. I am passionate about all things physics and am particularly keen on anything with a strong mathematical element. I also have a strong interest in the communication of science and maths to people of a wide range of abilities. I'm always looking for new ways to think about scientific concepts to make them more easily accessible. I believe that science and maths don't need to be as exclusive as they currently are!

During my time in the Air Cadets, I designed and delivered a teaching programme for teenagers of varying ages and abilities in order to teach subjects related to the principles of flight, history, navigation amongst other subjects. Whilst studying at university, I was also awarded the Forresters UoB Clear Direction prize for best scientific communicator. As a tutor, I feel that I can bring these skills together to make even the toughest scientific and mathematical concepts accessible in the lesson space.

I believe that everyone has their own unique learning style, particularly when it comes to subjects like maths and science. For this reason, I feel it is important that you get to choose the pace and style of the sessions. If, however, you feel unsure of where to start then we can begin with a general introduction to the area in question, slowly building up your confidence before we move onto questions, exam technique and problem-solving skills.

Initially I think it is important not to "learn an exam" so we will begin with broad approaches to topics in order to establish the basic concepts. From here we can begin looking at practise questions and how to apply exam technique in order to maximise marks in a currently challenging, and ever-changing, qualification system. On top of this, revision techniques are a huge part of the exam period and I'm keen not to let these be forgotten - I can always provide advice on different revision tactics.

As a rule, I think it’s important to enjoy the subjects you study in order to achieve in them. I hope that through a variety of teaching approaches and tasks it will be possible to improve not only your knowledge of a subject but your motivation and interest as well!

So, what's next? For any more information on the sessions or anything at all, don't hesitate to send me a message or book a 'Meet the Tutor' session. I look forward to meeting you soon!

I believe that everyone has their own unique learning style, particularly when it comes to subjects like maths and science. For this reason, I feel it is important that you get to choose the pace and style of the sessions. If, however, you feel unsure of where to start then we can begin with a general introduction to the area in question, slowly building up your confidence before we move onto questions, exam technique and problem-solving skills.

Initially I think it is important not to "learn an exam" so we will begin with broad approaches to topics in order to establish the basic concepts. From here we can begin looking at practise questions and how to apply exam technique in order to maximise marks in a currently challenging, and ever-changing, qualification system. On top of this, revision techniques are a huge part of the exam period and I'm keen not to let these be forgotten - I can always provide advice on different revision tactics.

As a rule, I think it’s important to enjoy the subjects you study in order to achieve in them. I hope that through a variety of teaching approaches and tasks it will be possible to improve not only your knowledge of a subject but your motivation and interest as well!

So, what's next? For any more information on the sessions or anything at all, don't hesitate to send me a message or book a 'Meet the Tutor' session. I look forward to meeting you soon!

Enhanced DBS Check

24/06/20145from 51 customer reviews

Lorna (Parent)

June 15 2017

My daughter Kitty has received tutoring from Ben in Physics GCSE for several weeks, up to her exam in June. She now awaits her results. Ben has worked very hard to help her to understand the subject, as with moving schools several times in the past 3 years she has had no consistent tuition. This resulted in her having no interest in Physics and Ben has helped her to look forward to and enjoy her sessions with him. He is persistent and encouraging, with a friendly manner and explains things well. I would recommend Ben as a tutor to young people at all levels

Lorna (Parent)

November 26 2016

Good session again.

Valery (Parent)

November 27 2017

Valery (Parent)

October 18 2017

Modulus-argument form implies that we should express z in terms of its straight line distance from the origin and the angle this straight line would make with the x axis. This is expressed as z = r e^{i*theta} where r is the modulus and theta is the argument. We thus wish to express r and theta in terms of x and y.

This problem is best solved visually by considering an Argand diagram for the general complex number, z. In this way we can see that z is represented as a point x units along the x-axis and y units along the y-axis, forming a right angled triangle with the vertical and horizontal.

As with any right angled triangle, using Pythagoras, we can see that the length of the hypotenuse (I.e. the distance from the origin, the modulus) is given by r = sqrt(x^{2}+y^{2}). Similarly, using basic trigonometry we can also see that the angle between this line and the x axis (theta, the argument) is given by theta = arctan(y/x).

This means that in modulus-argument form:

z = sqrt(x^{2}+y^{2}) exp(i*arctan(y/x))

Modulus-argument form implies that we should express z in terms of its straight line distance from the origin and the angle this straight line would make with the x axis. This is expressed as z = r e^{i*theta} where r is the modulus and theta is the argument. We thus wish to express r and theta in terms of x and y.

This problem is best solved visually by considering an Argand diagram for the general complex number, z. In this way we can see that z is represented as a point x units along the x-axis and y units along the y-axis, forming a right angled triangle with the vertical and horizontal.

As with any right angled triangle, using Pythagoras, we can see that the length of the hypotenuse (I.e. the distance from the origin, the modulus) is given by r = sqrt(x^{2}+y^{2}). Similarly, using basic trigonometry we can also see that the angle between this line and the x axis (theta, the argument) is given by theta = arctan(y/x).

This means that in modulus-argument form:

z = sqrt(x^{2}+y^{2}) exp(i*arctan(y/x))

While this question may appear unfamiliar at first, it is just a quadratic equation in disguise. Rearranging such that it equals 0:

x^2 + 5x - 36 = 0

Now this equation is ready to be solved! We can approach this in a couple of ways. I'll go through the factorisation method and using the quadratic formula.

__Factorisation Method__

This method is often the quickest way but requires a bit of practise to be able to spot the solution. If we look at a general factorisation where the coefficient of x is 1 in both brackets, i.e.

(x+a)(x+b) = 0

We can see when we expand this, we get

x^2 + ax + bx + ab = 0

x^2 + (a+b)x + ab = 0

Going back to our example, this shows us that we need to think of numbers such that a + b = 5 and ab = -36. The tricky part is actually thinking of these numbers.

Because we know that ab is a negative number, we know either a or b must be negative because a negative times a positive is a negative. We now need to trial and error a few potential numbers to find our answers. With a bit of good guessing we can see that a = -4 and b = 9 will work because -4x9 = -36 and -4+9 = 5.

This gives us

(x-4)(x+9)=0

For this to equal zero, one of these brackets must equal 0. If the first one equals 0 then,

x-4 = 0

x=4

or the second one,

x+9=0

x=-9

So our answer is x=4 or x=-9.

If you found it difficult to guess the right numbers then keep practising but, in the meantime, the formula method always works without guessing but does take a little longer...

__Formula Method__

For a general quadration equation ax^2 + bx + c = 0, the quadratic formula is given by

x = (-b +/- sqrt(b^2 - 4ac))/2a

For our example, a = 1, b = 5 and c = -36, so:

x = (-5 +/- sqrt(25 + (4x36)))/2

x = (-5 +/- sqrt(25+144))/2

x = (-5 +/- sqrt(169))/2

x = (-5 +/- 13)/2

Taking the + option:

x = (-5 + 13)/2

x = 8/2

x = 4

Taking the - option:

x = (-5-13)/2

x = -18/2

x = -9

Therefore, x=4 or x=-9 as expected from our first method.

While this question may appear unfamiliar at first, it is just a quadratic equation in disguise. Rearranging such that it equals 0:

x^2 + 5x - 36 = 0

Now this equation is ready to be solved! We can approach this in a couple of ways. I'll go through the factorisation method and using the quadratic formula.

__Factorisation Method__

This method is often the quickest way but requires a bit of practise to be able to spot the solution. If we look at a general factorisation where the coefficient of x is 1 in both brackets, i.e.

(x+a)(x+b) = 0

We can see when we expand this, we get

x^2 + ax + bx + ab = 0

x^2 + (a+b)x + ab = 0

Going back to our example, this shows us that we need to think of numbers such that a + b = 5 and ab = -36. The tricky part is actually thinking of these numbers.

Because we know that ab is a negative number, we know either a or b must be negative because a negative times a positive is a negative. We now need to trial and error a few potential numbers to find our answers. With a bit of good guessing we can see that a = -4 and b = 9 will work because -4x9 = -36 and -4+9 = 5.

This gives us

(x-4)(x+9)=0

For this to equal zero, one of these brackets must equal 0. If the first one equals 0 then,

x-4 = 0

x=4

or the second one,

x+9=0

x=-9

So our answer is x=4 or x=-9.

If you found it difficult to guess the right numbers then keep practising but, in the meantime, the formula method always works without guessing but does take a little longer...

__Formula Method__

For a general quadration equation ax^2 + bx + c = 0, the quadratic formula is given by

x = (-b +/- sqrt(b^2 - 4ac))/2a

For our example, a = 1, b = 5 and c = -36, so:

x = (-5 +/- sqrt(25 + (4x36)))/2

x = (-5 +/- sqrt(25+144))/2

x = (-5 +/- sqrt(169))/2

x = (-5 +/- 13)/2

Taking the + option:

x = (-5 + 13)/2

x = 8/2

x = 4

Taking the - option:

x = (-5-13)/2

x = -18/2

x = -9

Therefore, x=4 or x=-9 as expected from our first method.

The photoelectric effect is the emission of electrons from a metal when light is incident on the surface. The effect is due to the absorption of energy by the surface electrons causing them to become excited and thus released from the metal surface.

The wave theory for light would suggest then that, as long as we were to wait long enough, light source of any frequency should eventually cause the emission of electrons. It would also suggest that the kinetic energy of the emitted electrons would be controlled by the intensity of the light because, you would think, the brighter the light, the more energy it is providing.

However, it was observed that light intensity had no impact on the kinetic energy of the electrons and also that lower frequency light would not emit any electrons regardless of how long we wait.

On the other hand, the particle theory of light suggests that light is transmitted in 'packets' of energy called photons, each of energy E=hf, where f is the frequency and h is Planck's constant. It also suggests that electrons would absorb these photons on a one-on-one basis. This theory perfectly describes the observation as it suggests that unless an individual photon has enough energy, an electron will not be released from a metal. It also means that it doesn't matter how long we wait, if the photon doesn't have enough energy, the electrons will never be emitted. Since photon energy was only dependent on frequency, this would also explain why the electron kinetic energy was dependent on frequency and not the light intensity.

Other examples include the instantaneous emission of electrons and considerations of the work function of the material.

The photoelectric effect is the emission of electrons from a metal when light is incident on the surface. The effect is due to the absorption of energy by the surface electrons causing them to become excited and thus released from the metal surface.

The wave theory for light would suggest then that, as long as we were to wait long enough, light source of any frequency should eventually cause the emission of electrons. It would also suggest that the kinetic energy of the emitted electrons would be controlled by the intensity of the light because, you would think, the brighter the light, the more energy it is providing.

However, it was observed that light intensity had no impact on the kinetic energy of the electrons and also that lower frequency light would not emit any electrons regardless of how long we wait.

On the other hand, the particle theory of light suggests that light is transmitted in 'packets' of energy called photons, each of energy E=hf, where f is the frequency and h is Planck's constant. It also suggests that electrons would absorb these photons on a one-on-one basis. This theory perfectly describes the observation as it suggests that unless an individual photon has enough energy, an electron will not be released from a metal. It also means that it doesn't matter how long we wait, if the photon doesn't have enough energy, the electrons will never be emitted. Since photon energy was only dependent on frequency, this would also explain why the electron kinetic energy was dependent on frequency and not the light intensity.

Other examples include the instantaneous emission of electrons and considerations of the work function of the material.