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Two large parallel metal plates carry opposite charges. They are separated by 85mm. The work done by the field is 6x10-3J and its field exerts on a particle with charge +8µC. Calculate the surface charge density on each plate.

User Jhenrique
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1 Answer

27 votes
27 votes

Answer:

The surface charge density is
7.8*10^(-8) C/m^2.

Explanation:

separation, d = 85 mm

Work, W = 6 x 10^-3 J

charge , Q = 8µC

The potential difference is given by

W = q V


V=(6* 10^(-3))/(8* 10^(-6))=750 V

Let the charge on he capacitor is q.


q = CV\\\\q = (\varepsilon oA)/(d)* V\\\\(q)/(A) = (8.85* 10^(-12)*750)/(0.085) =7.8*10^(-8) C/m^2

User Pritam Barhate
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