10C and 20C are separated by a distance d. If the electric field at the location of a charge 10C is $\vec{E},$ the field at the location of 20 C is

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$\vec{E} / 2$

$- \vec{E} / 2$

$- \vec{E}$

$\vec{E}$

answer is B.

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Detailed Solution

E is due to $20 C; E = \frac{1}{4 {πε}_{0}} \frac{20}{r^{2}}$
E' is due to $10 C E' = \frac{1}{4 {πε}_{0}} \frac{10}{r^{2}} = \frac{- E}{2}$
(opposite hence negative)

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