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# Electric field vectors for two circular parallel plates

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Consider two circular parallel plates of radius R separated by a distance d &#61500;&#61500; R. Assume that these plates have uniformly distributed surface charges of Q and -Q. Electric field vectors in the region between the plates will all point perpendicular to the plates from the positive plate to the negative plate and will have a magnitude of 4&#61552;k&#61555;

a) Use equation d&#61542;=dVe/d=-qE-dr/q= d&#61542;=-E&#61655;dr
And the information just given to evaluate the potential difference between the plates in terms of Q, R, and d.
b) Argue that you get the same results using the formula for the potential of an infinite plate :
&#61542;-&#61542;s= 2&#61552;k&#61555; E=2&#61552;k&#61555;r

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The electric field vectors for two circular parallel plates is examined. The solution shows the formulas and calculations for arrive at the answer.

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a.)
Because,
Electric field strength is given as:
E = 4pi*k*s
here, s == sigma = Q/(pi*R^2)
k= 1/(4pi*e) (e stands for epsilon).
therefore,
E = ...

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###### Education
• MSc , Pune University, India
• PhD (IP), Pune University, India
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