Question

two particles A and B having charges of +2x10^-6 & - 4x10^-6 respectively are held fixed at a seperation of 20 cm. Locate the points on the line AB where (a) electric field is 0 (b) the electric potential is 0​

Answer 1

Since two charges are of opposite sign , so the point at which electric field is zero lies outside the system.

let x cm be the distance of the point from charge 2×10-6 C.

distance from the -4×10-6 C is (20+x) cm.

therefore,k×2×10-6x2 = k×4×10-6(20+x)2we getx = 20(1±2)since point lies outside the system , we take only positive sign.x = 20(1+2) cm

Let d be the distance from the charge 2×10-6 C at which potential is zero. Since potential is scalar quantity , this point lies inside the system.

k×2×10-6d + k×-4×10-620-d = 0we getd = 20/3 cm

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