JEE Advance - Physics (2018 - Paper 2 Offline - No. 14)

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

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

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

$$P \to 4;Q \to 3;R \to 1,2;S \to 4$$

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

Explanation

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 $$R \to 1$$.

For a small dipole with point charges Q at (0, 0, l) and $$-$$Q at (0, 0, $$-$$l), 2l << d, we have

$$E = {1 \over {4\pi {\varepsilon _0}}}{{2Q \times 2l} \over {{d^3}}}$$

$$ \Rightarrow E \propto {1 \over {{d^3}}}$$

Thus, the correct mapping is $$S \to 2$$.

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 $$Q \to 3$$.

For two infinite wires carrying uniform linear charge density parallel to the x-axis, we have

$$E = {\lambda \over {2\pi {\varepsilon _0}(d - l)}} - {\lambda \over {2\pi {\varepsilon _0}(d + l)}} = {\lambda \over {2\pi {\varepsilon _0}}}\left( {{1 \over {d - l}} - {1 \over {d + l}}} \right)$$

$$ \Rightarrow E = {\lambda \over {2\pi {\varepsilon _0}}}\left( {{{d + l - d + l} \over {(d - l)(d + l)}}} \right) = {\lambda \over {2\pi {\varepsilon _0}}}{{(2l)} \over {({d^2} - {l^2})}}$$

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 $$R \to 4$$.

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 $$P \to 5$$.

Therefore, option (B) is correct.

JEE Advance - Physics (2018 - Paper 2 Offline - No. 14)

Explanation

Comments (0)