Question

Two parallel plates have a potential difference of 10V between them.If the plates are 0.5 mm apart,what will be the strength of electric charge.​

Answer 1

Answer:

E = -dV/dr

E = (- 10/0.5e-3) V/m

E = (-20e3) V/m

E = - 20,000 V/m

Answer 2

Electric field strength, E = 20000 N/C

Explanation:

Given that,

Potential difference between two parallel plates, V = 10 V

Separation between the plates, d = 0.5 mm

The relation between the electric field and electric potential is given by :

$E=\dfrac{V}{d}$

$E=\dfrac{10\ V}{0.5\times 10^{-3}\ m}$

E = 20000 N/C

So, the electric field strength is 20000 N/C. Hence, this is the required solution.

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Topic : electric field

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