JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 13)
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 :-
$$E \propto {1 \over D^3}$$
$$E \propto {1 \over D}$$
$$E \propto {1 \over D^4}$$
$$E \propto {1 \over D^2}$$
Explanation
Electric field at p = 2E1cos$$\theta $$1 –2E2cos$$\theta $$2
= $${{2Kq} \over {\left( {{d^2} + {D^2}} \right)}} \times {D \over {{{\left( {{d^2} + {D^2}} \right)}^{1/2}}}} - {{2Kq} \over {\left[ {{{\left( {2d} \right)}^2} + {D^2}} \right]}} \times {D \over {{{\left[ {{{\left( {2d} \right)}^2} + {D^2}} \right]}^{1/2}}}}$$
$$ = 2KqD\left[ {{{\left( {{d^2} + {D^2}} \right)}^{ - 3/2}} - {{\left( {4{d^2} + {D^2}} \right)}^{ - 3/2}}} \right]$$
$$ = {{2KqD} \over {{D^3}}}\left[ {{{\left( {1 + {{{d^2}} \over {{D^2}}}} \right)}^{ - 3/2}} - {{\left( {1 + {{4{d^2}} \over {{D^2}}}} \right)}^{ - 3/2}}} \right]$$
Applying binomial approximation $$ \because $$ d << D
$$ = {{2KqD} \over {{D^3}}}\left[ {1 - {3 \over 2}{{{d^2}} \over {{D^2}}} - \left( {1 - {{3 \times 4{d^2}} \over {2{D^2}}}} \right)} \right]$$
$$ = {{2KqD} \over {{D^3}}}\left[ {{{12} \over 2}{{{d^2}} \over {{D^2}}} - {3 \over 2}{{{d^2}} \over {{D^2}}}} \right]$$
$$ = {{9kq{d^2}} \over {{D^4}}}$$
= $${{2Kq} \over {\left( {{d^2} + {D^2}} \right)}} \times {D \over {{{\left( {{d^2} + {D^2}} \right)}^{1/2}}}} - {{2Kq} \over {\left[ {{{\left( {2d} \right)}^2} + {D^2}} \right]}} \times {D \over {{{\left[ {{{\left( {2d} \right)}^2} + {D^2}} \right]}^{1/2}}}}$$
$$ = 2KqD\left[ {{{\left( {{d^2} + {D^2}} \right)}^{ - 3/2}} - {{\left( {4{d^2} + {D^2}} \right)}^{ - 3/2}}} \right]$$
$$ = {{2KqD} \over {{D^3}}}\left[ {{{\left( {1 + {{{d^2}} \over {{D^2}}}} \right)}^{ - 3/2}} - {{\left( {1 + {{4{d^2}} \over {{D^2}}}} \right)}^{ - 3/2}}} \right]$$
Applying binomial approximation $$ \because $$ d << D
$$ = {{2KqD} \over {{D^3}}}\left[ {1 - {3 \over 2}{{{d^2}} \over {{D^2}}} - \left( {1 - {{3 \times 4{d^2}} \over {2{D^2}}}} \right)} \right]$$
$$ = {{2KqD} \over {{D^3}}}\left[ {{{12} \over 2}{{{d^2}} \over {{D^2}}} - {3 \over 2}{{{d^2}} \over {{D^2}}}} \right]$$
$$ = {{9kq{d^2}} \over {{D^4}}}$$
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