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Solution :

(a) zero at both the points <br> Charge -q is located at (0,0,-a) and charge +q is located at (0,0,a) . Hence they from a dipole. Point (0,0,z) is on the axis of this dipole and point (x,y,0) is normal to the axis of the dipole . Hence, electrostatic potential `(V_(1)) ` at point (5,0,0) is given by <br> `V_(1)=(-q)/(4piin_(0))1/(sqrt((5-0)^(2)+(-a)^(2)))+q/(4piin_(0))1/((5-0)^(2)+a^(2))` <br> `=(-q)/(4piin_(0)sqrt(25^(2)+a^(2))+q/(4pi in_(0)sqrt(25+a^(2))` <br> =0 <br> Electrostatic potentials, `V_(2)`, at point `(-7,0,0)` is given by <br> V_(2)=(-q)/(4pi in_(0))1/(sqrt(-7)^(2)+(-a)^(2))+q/(4pi in_(0))1/(sqrt((-7)^(2)+(a)^(2)))` <br> `=(-q)/(4pi in_(0))sqrt(49+a^(2))+q/(4pi in_(0))1/(sqrt(49+a^(2)))` <br> =0 <br> Hence, no work is done in moving a small test charge from point (5,0,0) to point (-7,0,0) along the x-axis . The answer does not change because work doen by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points. Related Video

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