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(B) The electric potential $V$ at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.From the figure,

Vedclass · Accepted Answer

$\left(\frac{3}{8}\right) qK$

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(B) The electric potential $V$ at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.From the figure,the distance of point $P$ from the charge $+q$ is $r_1 = 2 \ cm = 2 	imes 10^{-2} \ m$.The distance of point $P$ from the charge $-q$ is $r_2 = 3 \ cm + 3 \ cm + 2 \ cm = 8 \ cm = 8 	imes 10^{-2} \ m$.The potential at point $P$ is given by:$V = V_{+q} + V_{-q} = \frac{Kq}{r_1} + \frac{K(-q)}{r_2}$$V = Kq \left( \frac{1}{2 	imes 10^{-2}} - \frac{1}{8 	imes 10^{-2}} ight)$$V = Kq \left( \frac{4 - 1}{8 	imes 10^{-2}} ight) = Kq \left( \frac{3}{8 	imes 10^{-2}} ight)$$V = Kq \left( \frac{3}{8} ight) 	imes 10^2 \ V$.Thus,the potential in units of $10^2 \ V$ is $\left( \frac{3}{8} ight) qK$.

An electric dipole is placed as shown in the figure. The electric potential (in $10^2 \ V$) at point $P$ due to the dipole is ($\epsilon_0 =$ permittivity of free space and $\frac{1}{4 \pi \epsilon_0} = K$):

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