Two point charges of + 10 microcoulomb and + 40 microcoulomb respectively are placed 12 CM apart find the position of the point where electric field is zero

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Solution

Let the point where electric field is 0, be at a distance x from the first charge and (12-x)cm from the second charge.

then, electric field due to first charge is E1=910^91010^-6/x^2 =910^4/x^2

electric field due to second charge is E2=910^94010^-6/(12-x)^2 =3610^4/(12-x)^2

E1+E2=0 => E1=-E2

on solving the underlined equation, you will get the answer.

910^4/x^2 = 3610^4/(12-x)^2

taking root on both sides

310^2/x = 610^2/(12-x)
3/x = 6/(12-x)
1/x = 2/(12-x)
12-x = 2x
3x = 12
x = 4
4 cm from the first charge the E will be zero.

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