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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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