Force of interaction between fast moving capacitor plates.

Electricity and Magnetism Level 5

The plates of a square plate capacitor are separated by a very small distance and have charge $q = 1 \mu C$ side length $l = 1m$ suddenly start moving in opposite directions with equal speed $v = 2 \times 10^8 m/s$ as shown. Find the force experienced by one plate due to the other at this instant in Newtons .

The answer is 0.0815.

1 solution

Jatin Yadav
Jun 19, 2014

We know that the electric force between plates of a parallel plate capacitor is $F_{E} = \dfrac{Q^2}{2A \epsilon_{0}} = \dfrac{Q^2}{2L^2 \epsilon_{0}}$

Now, we know that motion of charges creates magnetic field. Now, due to the magnetic field due to motion of charges of left plate, charges of right plate will experience magnetic force.

Effective current flowing = $i = \dfrac{dq}{dt} = \dfrac{\sigma L dx}{dt} = \dfrac{Qv}{L}$

Now, We use ampere's law to find magnetic field in region close to the plate.

Consider an amperian square loop of side $x$ extending equally on both sides of left plate.

Clearly, $B \times 2x = \mu_{0} i \dfrac{x}{L}$

$\Rightarrow B = \dfrac{\mu_{0} Q v}{2L^2}$

Hence, $F_{B} = Q v B = \dfrac{\mu_{0} Q^2 v^2}{2 L^2} = F_{E} \dfrac{v^2}{c^2}$

Note that both the forces would be attracting. Hence, they add to each other.

Hence, , $F = F_{E} \bigg( 1 + \dfrac{v^2}{c^2}\bigg)$

Put values to get the answer as $8.17$

Note : it has been assumed that the distance between plates is very small. This is because the plates of a capacitor are very close to each other.

I think fringing E field would also have significant effects.

Pranay Pratyush - 6 years, 11 months ago

Good question. I just think you could have written some assumptions, such as: consider the distance between the plates too small, don't consider the industion between them...

Felipe Hofmann - 6 years, 10 months ago

Interesting problem. I didn't get the correct answer because I had this intuition that moving plates (in sideways directions) would experience a lower force than if it were static. So I thought the answer was ${F} = {F}_{E} \left(1-{\left(\frac{v}{c}\right)}^{2}\right)$ .

Steven Zheng - 6 years, 10 months ago

wrong solution fringing forces cannot be neglected

hemant khatri - 6 years, 9 months ago

Hi, the distance between the plates is very small, and the fringing force would be negligible

jatin yadav - 6 years, 8 months ago

Where am I going wrong? $F={ \frac { { Q^{ 2 } } }{ 2L^{ 2 }\epsilon _{ 0 } } }(1+{ \frac { { v^{ 2 } } }{ c^{ 2 } } })\\ F={ \frac { { 10^{ -12 } } }{ 2(1)^{ 2 }(8.854\times 10^{ -12 }) } }(1+{ \frac { { 4\times 10^{ 16 } } }{ 9\times 10^{ 16 } } })\\ F=\frac { 13 }{ 18\times 8.854 } =0.0815$

Dhruva Patil - 6 years, 5 months ago

I am also getting 0.081.. done exactly what you have done..!!

Rohit Gupta - 6 years ago

Dhruva Patil ; I am also getting the same answer as you.

Karan Siwach - 6 years, 4 months ago

@Karan Siwach Guess they made some calculation error.

Dhruva Patil - 6 years, 4 months ago

You are right . I am sorry for the inconvenience caused. It has been edited now. Thanks :)

jatin yadav - 6 years ago

I think the answer was correct earlier-8.16 In the calculation above, length has been taken 1m while in question it is 0.1m which multiplies the answer by factor of 100, hence 8.16 . Did Exactly what you did but answer is 8.16 . Do please check and correct

Akul Agrawal - 5 years, 11 months ago

length is 0.1m not 1m hence ans. must be 8.16

Akul Agrawal - 5 years, 11 months ago

No consideration of relativity here? :P Just saying.

A Former Brilliant Member - 6 years, 1 month ago

Force of interaction between fast moving capacitor plates.

The answer is 0.0815.

1 solution

0 pending reports