Tackle the capacitors!

Electricity and Magnetism Level 4

Two identical capacitors having plate separation $d_0$ are connected parallel to each other across points $A$ and $B$ as shown.A charge Q is imparted to the system by connecting a battery across $A$ and $B$ and the battery is removed.Now the first plate of the first capacitor and second plate of the second capacitor starts moving with constant velocity $u_0$ towards left.Then the magnitude of current flowing in the loop if $Q=3 C;u_0=4 m/s;d_0=1m$ is $?$

The answer is 6.00.

1 solution

Rajdeep Brahma
Jul 2, 2018

Let each plate move a distance $x$ from its initial position.Let $q$ charge flow in the loop.

So $\frac{(Q/2-q)(d+x)}{\epsilonA}$ - $\frac{(Q/2+q)(d-x)}{\epsilonA}$ $=0$

q= $\frac{Qx}{2d_0}$ and so I= $\frac{dq}{dt}=\frac{Qu_0}{2d_0}$ =6 A

Isn't it the current in the loop initially?

Md Zuhair - 2 years, 11 months ago

It is It also happens to be a constant current Due to the fact that the transferrence of charges occurs intermittently between the 2 capacitors (We neglect the transient state or end behaviours)

Suhas Sheikh - 2 years, 11 months ago

Yeah right

rajdeep brahma - 2 years, 11 months ago

Tackle the capacitors!

The answer is 6.00.

1 solution

Let each plate move a distance x x x from its initial position.Let q q q charge flow in the loop.

So ( Q / 2 − q ) ( d + x ) ϵ ∗ A \frac{(Q/2-q)(d+x)}{\epsilon*A} ϵ ∗ A ( Q / 2 − q ) ( d + x ) ​ - ( Q / 2 + q ) ( d − x ) ϵ ∗ A \frac{(Q/2+q)(d-x)}{\epsilon*A} ϵ ∗ A ( Q / 2 + q ) ( d − x ) ​ = 0 =0 = 0

q= Q ∗ x 2 ∗ d 0 \frac{Q*x}{2*d_0} 2 ∗ d 0 ​ Q ∗ x ​ and so I= d q d t = Q ∗ u 0 2 ∗ d 0 \frac{dq}{dt}=\frac{Q*u_0}{2*d_0} d t d q ​ = 2 ∗ d 0 ​ Q ∗ u 0 ​ ​ =6 A

0 pending reports

Let each plate move a distance $x$ from its initial position.Let $q$ charge flow in the loop.

So $\frac{(Q/2-q)(d+x)}{\epsilonA}$ - $\frac{(Q/2+q)(d-x)}{\epsilonA}$ $=0$

q= $\frac{Qx}{2d_0}$ and so I= $\frac{dq}{dt}=\frac{Qu_0}{2d_0}$ =6 A