An electric dipole is placed as shown in the figure. The electric potential (in 10² V) at point P due to the dipole is ( ε₀ = permittivity of free space and 1/4πε₀=K ): 1. (8/3)qK 2. (3/8)qK 3. (5/8)qK 4. (8/5)qK

Answer: 2. (3/8)qK P lies on the axis of the dipole. Distances from P: to +q it is 5-3=2 cm, to -q it is 5+3=8 cm. V=Kq(dfrac10.02-dfrac10.08)=Kq(50-12.5)=37.5 Kq=3/8×10² Kq volt. In units of 10² V the potential is 3/8 qK .

Topic: Electric potential, potential difference, equipotential surfaces· Electrostatics

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 (ε_0 = permittivity of free space and 1/4πε_0=K):

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 \(\frac{1}{4\pi\epsilon_0}=K\)):

-q and +q 3 cm either side of O, point P 5 cm from O

1. \(\left(\frac{8}{3}\right)qK\)

2. \(\left(\frac{3}{8}\right)qK\)

3. \(\left(\frac{5}{8}\right)qK\)

4. \(\left(\frac{8}{5}\right)qK\)

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Answer: 2. \(\left(\frac{3}{8}\right)qK\)

P lies on the axis of the dipole. Distances from P: to \(+q\) it is \(5-3=2\) cm, to \(-q\) it is \(5+3=8\) cm.

\(V=Kq\left(\dfrac1{0.02}-\dfrac1{0.08}\right)=Kq(50-12.5)=37.5\,Kq=\dfrac38\times10^2\,Kq\) volt.

In units of \(10^2\) V the potential is \(\dfrac38\,qK\).

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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 (ε_0 = permittivity of free space and 1/4πε_0=K):

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