Question 777: Electrostatics & Capacitance

Question

Identical charges ($$-$$q) are placed at each corners of cube of side b then electrostatic potential energy of charge (+q) which is placed at centre of cube will be

Accepted Answer

There are eight corners of a cube and in each corner there is a charge of (Ã¢ÂÂq). At the centre of the corner there is a charge of (+q). Each corner is equidistant from the centres of the cube and the distance (d) is half of the diagonals of the cube.
Diagonal of the cube = $$\sqrt {{b^2} + {b^2} + {b^2}}  = \sqrt 3 b$$
$$ 	herefore $$ $$d = \sqrt 3 b/2$$
Now, electric potential energy of the charge (+q) due to a charge (Ã¢ÂÂq) at one corner
U $$ = {{{q_1}{q_2}} \over {4\pi {\varepsilon _0}r}} = {{\left( { + q} ight) 	imes \left( { - q} ight)} \over {4\pi {\varepsilon _0}\left( {\sqrt 3 b/2} ight)}} =  - {{{q^2}} \over {2\pi {\varepsilon _0}\left( {\sqrt 3 b} ight)}}$$
$$ 	herefore $$ Total electric potential energy due to all the eight identical charges
= $$8U =  - {{8{q^2}} \over {2\pi {\varepsilon _0}\sqrt 3 b}} = {{ - 4{q^2}} \over {\sqrt 3 \pi {\varepsilon _0}b}}$$

Answer

$${{ - 4\sqrt 2 {q^2}} \over {\pi {\varepsilon _0}b}}$$

Answer

$${{ - 8\sqrt 2 {q^2}} \over {\pi {\varepsilon _0}b}}$$

Answer

$${{ - 4\,{q^2}} \over {\sqrt 3 \,\pi {\varepsilon _0}b}}$$

Answer

$${{8\sqrt 2 \,{q^2}} \over {4\,\pi {\varepsilon _0}b}}$$

Electrostatics & Capacitance

Identical charges ($-$q) are placed at each corners of cube of side b then electrostatic potential energy of charge (+q) which is placed at centre of cube will be

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