Question 571: Electrostats & Capacitance

Question

Four point charges –q, +q, +q and –q are placed
on y-axis at y = –2d, y = –d, y = +d and
y = +2d, respectively. The magnitude of the
electric field E at a point on the x-axis at
x = D, with D >> d, will behave as :-

Accepted Answer

Electric field at p = 2E
1
cos$$	heta $$
1
Ã¢ÂÂ2E
2
cos$$	heta $$
2
= $${{2Kq} \over {\left( {{d^2} + {D^2}} ight)}} 	imes {D \over {{{\left( {{d^2} + {D^2}} ight)}^{1/2}}}} - {{2Kq} \over {\left[ {{{\left( {2d} ight)}^2} + {D^2}} ight]}} 	imes {D \over {{{\left[ {{{\left( {2d} ight)}^2} + {D^2}} ight]}^{1/2}}}}$$
$$ = 2KqD\left[ {{{\left( {{d^2} + {D^2}} ight)}^{ - 3/2}} - {{\left( {4{d^2} + {D^2}} ight)}^{ - 3/2}}} ight]$$
$$ = {{2KqD} \over {{D^3}}}\left[ {{{\left( {1 + {{{d^2}} \over {{D^2}}}} ight)}^{ - 3/2}} - {{\left( {1 + {{4{d^2}} \over {{D^2}}}} ight)}^{ - 3/2}}} ight]$$
Applying binomial approximation $$ \because $$ d << D
$$ = {{2KqD} \over {{D^3}}}\left[ {1 - {3 \over 2}{{{d^2}} \over {{D^2}}} - \left( {1 - {{3 	imes 4{d^2}} \over {2{D^2}}}} ight)} ight]$$
$$ = {{2KqD} \over {{D^3}}}\left[ {{{12} \over 2}{{{d^2}} \over {{D^2}}} - {3 \over 2}{{{d^2}} \over {{D^2}}}} ight]$$
$$ = {{9kq{d^2}} \over {{D^4}}}$$

Answer

$$E \propto {1 \over D^3}$$

Answer

$$E \propto {1 \over D}$$

Answer

$$E \propto {1 \over D^4}$$

Answer

$$E \propto {1 \over D^2}$$

Electrostats & Capacitance

Four point charges –q, +q, +q and –q are placed on y-axis at y = –2d, y = –d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D >> d, will behave as :-

Choose the correct answer: