Q: Calculate electric field due to dipole of dipole momentum P at distance r and angle \theta .

E_1 = \dfrac{1}{4\pi \varepsilon _0 } \times \dfrac{2p \cos \theta}{r^3} (along OC) .... (i)Simlialry, P lies on the equatorial line of dipole with moment \vec{p}_2 . Hence, the magnitude of electric field intensity ]vec{E}_2 at P due to \vec{p}_2 is E_2 = \dfrac{1}{4\pi \varepsilon _0} - \dfrac{p \sin \theta}{r^3} (opposite to p_2 ) ...(ii)Hence, the resultant intensity at P is \vec{E} = \vec{E}_1 + \vec{E}_2 .The magnitude of \vec{E} is E = \sqrt{(E^2_1 + E^2_2)} (as \vec{E}_1 and \vec{E}_2 are mutally perpendicular).or E = \sqrt{\left(\dfrac{2p\cos \theta}{4\pi \varepsilon _0 r^3}\right)^2 + \left(\dfrac{p \sin \theta}{4\pi \varepsilon _0 r^3}\right)^2} = \dfrac{p}{4\pi \varepsilon _0 r^3} \sqrt{4\cos^2 \theta + \sin^2 \theta} = \dfrac{p}{4\pi \varepsilon _0 r^3} \sqrt{1 +3 \cos^2 \theta}

Question 1

Charge Q is distributed uniformly on length  l  of a wire. It is bent in the form of a semicircular ring. Find the electric field at the centre of the ring.

Accepted Answer

Consider two small elements of length  dl  each charge  dq  on each element. If  \lambda  is the line charge density,  Q=\lambda l=\lambda . \pi r . as wire is bent in the form of semicircular ring.Hence ,    dq=\lambda dl=\frac{Q}{\pi r}.(r d	heta)=\frac{Q}{\pi} d	heta  .  Here sine component of the field at center cancel out and  only cosine will contribute the filed.thus,  \displaystyle dE_c=2dE\cos	heta=2\frac{dq}{4\pi\epsilon_0 r^2}\cos	heta=\frac{Q}{2\pi^2 \epsilon_0 r^2}\cos	heta d	heta  or  \displaystyle E_c=\frac{Q}{2\epsilon_0(\pi r)^2}\int_0^{\pi/2}\cos	heta d	heta=\frac{Q}{2\epsilon_0(\pi r)^2}(\sin\frac{\pi}{2}-\sin0)=\frac{Q}{2\epsilon_0l^2}      as  (l=\pi r)

Question 2

Suppose we have a uniformly charged circular ring of radius

R

and charge

Q

. The electric field at a distance

x

from the centre such that

x > > R

is given by :

Take

k = \frac{1}{4 π ϵ _{0}}

.

Accepted Answer

Let us consider a small charge element of charge  dq And,  dq=\dfrac{Q}{2\pi R} .The field at point P due to this element is  =E=\dfrac{Kdq}{ r^2} E=\dfrac{Kdq}{ (R^2+x^2)} Now, from figure we see that component of field normal to axis is cancelled by two diametrically opposite points.Hence, only component of field along axis is left which add up for all such elements. E_{net}=\displaystyle\int E\cos	heta      where  	heta  is same for all elements means  	heta=constant \implies E_{net}=\displaystyle\int\dfrac{K\cos	heta}{(R^2+x^2)}dq \implies E_{net}=\dfrac{KQ}{(R^2+x^2)}\cos	heta \implies E_{net}=\dfrac{KQ}{(R^2+x^2)}\dfrac{x}{\sqrt{R^2+x^2}} \implies E_{net}=\dfrac{KQx}{(R^2+x^2)^{3/2}} For  x>>R ,we can say  x^2+R^2\approx x^2 E_{net}=\dfrac{KQx}{(x^2)^{3/2}}=\dfrac{KQx}{x^3} \implies E_{net}=\dfrac{KQ}{x^2} Answer-(B)

Question 3

Calculate electric field due to dipole of dipole momentum  P  at distance  r  and angle  	heta .

Accepted Answer

E_1 = \dfrac{1}{4\pi \varepsilon _0 } 	imes \dfrac{2p \cos 	heta}{r^3}    (along OC)     .... (i)Simlialry,  P  lies on the equatorial line of dipole with moment  \vec{p}_2 . Hence, the magnitude of electric field intensity  ]vec{E}_2  at  P  due to  \vec{p}_2  is E_2 = \dfrac{1}{4\pi \varepsilon _0} - \dfrac{p \sin 	heta}{r^3}    (opposite to  p_2 )       ...(ii)Hence, the resultant intensity at  P  is  \vec{E} = \vec{E}_1 + \vec{E}_2 .The magnitude of  \vec{E}  is  E = \sqrt{(E^2_1 + E^2_2)}    (as  \vec{E}_1  and  \vec{E}_2  are mutally perpendicular).or   E = \sqrt{\left(\dfrac{2p\cos 	heta}{4\pi \varepsilon _0 r^3}ight)^2 + \left(\dfrac{p \sin 	heta}{4\pi \varepsilon _0 r^3}ight)^2} = \dfrac{p}{4\pi \varepsilon _0 r^3} \sqrt{4\cos^2 	heta + \sin^2 	heta} = \dfrac{p}{4\pi \varepsilon _0 r^3} \sqrt{1 +3 \cos^2 	heta}

Question 4

A short dipole of dipole moment P is placed near a point charge Q as shown in figure. Find force on the dipole due to the point charge

Accepted Answer

Force on the point charge due to the dipole F = (Q)E_{dipole} F = (Q) \left(\dfrac{2KP}{r^3}\right)  (towards right)For action-reaction concept, force on the dipole due to point charge will be equal to the force on charge due to dipole.Hence, force on dipole, F' = \dfrac{2KPQ}{r^3}  (towards left)

Electric Charges And Fields