In the figure below, the electric field lines

CBSE Class 12 CBSE Class 12 Physics
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1

In the figure below, the electric field lines on the left have twice the separation of those on the right, (i) If the magnitude of the field of A is 40 N/Q then what force acts on a proton at A? (ii) What is the magnitude, of the field at B?

2

(i)Charge of proton , q = 1.6 x {{10}^{-19}} C
Force on proton at A is F = q { E }_{ A }
= 1.6 x {{10}^{-19}} x 40N/C
= 6.4 x {{10}^{-18}} N
(ii) Since, electric field,

Hence, { E }_{ B } = 1/2 { E }_{ A }
= 1/2(40N/C)
= 20 N/C

3

But electric field is related to 1/r^2 so it should get 1/4 times ???