Question 145: Electrostats & Capacitance

Question

The electric field $$E$$ is measured at a point $$P(0,0,d)$$ generated due to various charge distributions and the dependence of $$E$$ on $$d$$ is found to be different for different charge distributions. List-$${\rm I}$$  contains different relations between $$E$$ and $$d$$. List-$${\rm II}$$ describes different electric charge distributions, along with their locations. Match the functions in List-$${\rm I}$$ with the related charge distributions in List-$${\rm II}$$.

LIST - I 
LIST - II

P.
$$E$$ is independent of $$d$$
1.
A point charge Q at the origin

Q.
$$E\, \propto \,1/d$$
2.
A small dipole with point charges $$Q$$ at $$\left( {0,0,l} \right)$$ and $$-Q$$ at $$\left( {0,0, - l} \right).$$ Take $$2l <  < d$$

R.
$$E\, \propto \,1/{d^2}$$
3.
An infinite line charge coincident  with the x-axis, with uniform linear charge density $$\lambda $$

S.
$$E\, \propto \,1/{d^3}$$
4.
Two infinite wires carrying uniform linear charge density parallel to the $$x$$-axis. The one along $$\left( {y = 0,z = l} \right)$$ has a charge density $$ + \lambda $$ and the one 
along $$\left( {y = 0,z =  - l} \right)$$ has a charge density Take

5.
Infinite plane charge coincident 
with the $$xy$$-plane with uniform surface charge density

Accepted Answer

For a point charge Q at the origin, we have
$$E = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{d^2}}}$$
$$ \Rightarrow E \propto {1 \over {{d^2}}}$$
Thus, the correct mapping is $$R 	o 1$$.
For a small dipole with point charges Q at (0, 0, l) and $$-$$Q at (0, 0, $$-$$l), 2l << d, we have
$$E = {1 \over {4\pi {\varepsilon _0}}}{{2Q 	imes 2l} \over {{d^3}}}$$
$$ \Rightarrow E \propto {1 \over {{d^3}}}$$
Thus, the correct mapping is $$S 	o 2$$.
For an infinite line charge coincident with x-axis with uniform linear charge density $$\lambda$$, we have
$$E = {\lambda  \over {2\pi {\varepsilon _0}d}}$$
$$ \Rightarrow E \propto {1 \over d}$$
Thus, the correct mapping is $$Q 	o 3$$.
For two infinite wires carrying uniform linear charge density parallel to the x-axis, we have
$$E = {\lambda  \over {2\pi {\varepsilon _0}(d - l)}} - {\lambda  \over {2\pi {\varepsilon _0}(d + l)}} = {\lambda  \over {2\pi {\varepsilon _0}}}\left( {{1 \over {d - l}} - {1 \over {d + l}}} ight)$$
$$ \Rightarrow E = {\lambda  \over {2\pi {\varepsilon _0}}}\left( {{{d + l - d + l} \over {(d - l)(d + l)}}} ight) = {\lambda  \over {2\pi {\varepsilon _0}}}{{(2l)} \over {({d^2} - {l^2})}}$$
Since $$2l <  < d$$
$$ \Rightarrow E = {\lambda  \over {2\pi {\varepsilon _0}}}{{2l} \over {{d^2}}}$$
$$ \Rightarrow E \propto {1 \over {{d^2}}}$$
Thus, the correct mapping is $$R 	o 4$$.
For an infinite plane charge coincident with the x-y plane with uniform surface charge density, we have
$$E = {\sigma  \over {2\pi {\varepsilon _0}}}$$
Therefore, E is independent of d.
Thus, the correct mapping is $$P 	o 5$$.
Therefore, option (B) is correct.

Answer

$$P 	o 5;Q 	o 3,4;R 	o 1;S 	o 2$$

Answer

$$P 	o 5;Q 	o 3;R 	o 1,4;S 	o 2$$

Answer

$$P 	o 4;Q 	o 3;R 	o 1,2;S 	o 4$$

Answer

$$P 	o 4;Q 	o 2,3;R 	o 1;S 	o 5$$

Electrostats & Capacitance

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