Question 717: Electrostatics & Capacitance

Question

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 ($$\in_0$$ = permittivity of free space and $$\frac{1}{4 \pi \epsilon_{0}}$$ = K) :

Accepted Answer

$$\mathrm{v}=\frac{\mathrm{Kq}}{2 	imes 10^{-2}}-\frac{\mathrm{Kq}}{8 	imes 10^{-2}}$$
$$=\mathrm{Kq}\left[\frac{3}{8}ight] 	imes 10^{2}$$

Answer

$$\left(\frac{5}{8}\right) \mathrm{qK}$$

Answer

$$\left(\frac{8}{5}\right) \mathrm{qK}$$

Answer

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

Answer

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

Electrostatics & Capacitance

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 ($\in_0 = permittivity of free space and \frac{1}{4 \pi \epsilon_{0}}$ = K) :

Choose the correct answer: