- #1

athrun200

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## Homework Statement

The problem I would like to ask is in Griffiths Introduction to Electrodynamics p358 problem 8.6 part (c)

## Homework Equations

## The Attempt at a Solution

The answer is $${\varepsilon _0}EBAd\hat y$$ which is same as part (a)

Here's how I found a loophole:

One key information that will come in handy later is: the magnetic field is now decreasing, which will produce a electric field.

Before finding the impulse, we need to find the force acting on the capacitor since

$$\overrightarrow I = \int {\overrightarrow F \cdot dt} $$

The force, which is caused by the induced electric field, is simply

$$\overrightarrow F = \sigma A[ - E(d) + E(0)]\widehat y$$

To eliminate the induced electric field, I use the following equation

$$\oint {\vec E \cdot d\vec l} = \int\limits_S {\vec \nabla \times \vec E \cdot d\vec a = - ld{{\partial \vec B} \over {\partial t}}} $$

Now consider only the line integral on the left hand side

$$\oint {\vec E \cdot d\vec l} = \int\limits_l^0 {E(d)dx + } \int\limits_d^0 {{E_1}dy + } \int\limits_0^l {E(0)dx + } \int\limits_0^d {{E_2}dl} $$

$$ = l[ - E(d) + E(0)] + d({E_2} - {E_1})$$

Here's the problem. Useless $${E_2} - {E_1} = 0$$

,otherwise there is no way to find $$ - E(d) + E(0)$$

However $${E_2} - {E_1} = 0$$ implies $${E_2} = {E_1}$$

this symmetry would cause $$ E(d) = E(0)$$ otherwise it is really awkward.

What's wrong?

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