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Q.Find the electric potential at any point due to an electric dipole. OR Obtain the capacitance of a parallel plate capacitor. What is the function of a dielectric in a capacitor?

Nagaland NbseNagaland Board of School Education 2024Subjective· 3mImportance★★★★★
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Superposing the potentials of the +q and −q charges of a dipole and expanding for r≫ar\gg a gives V=pcos⁡θ4πϵ0r2V=\dfrac{p\cos\theta}{4\pi\epsilon_0 r^2}.

Consider a dipole with charges +q+q at A and −q-q at B, separated by distance 2a2a, with dipole moment p=q(2a)p=q(2a) directed from −q-q to +q+q. Let P be a point at distance rr from the centre O, with the line OP making angle θ\theta with the dipole axis.

Let r1r_1 = distance from +q+q to P, and r2r_2 = distance from −q-q to P. For r≫ar\gg a, using the geometry (dropping perpendiculars from A and B onto OP):

r1≈r−acos⁡θ,r2≈r+acos⁡θr_1 \approx r - a\cos\theta, \qquad r_2 \approx r + a\cos\theta

The net potential at P (scalar sum of the two point-charge potentials):

V=14πϵ0(qr1−qr2)=q4πϵ0⋅r2−r1r1r2V = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{r_1}-\frac{q}{r_2}\right) = \frac{q}{4\pi\epsilon_0}\cdot\frac{r_2-r_1}{r_1r_2}

Now r2−r1≈2acos⁡θr_2-r_1 \approx 2a\cos\theta, and for r≫ar\gg a, r1r2≈r2r_1r_2\approx r^2. So: …

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