Question

A test charge q is moved without acceleration from A to C along the path from A to B and then from B to C in electric field as shown:
a) Calculate potential difference between A to C.
b) At which point (of the two) is the electric potential more.

Answer 1

(a) E = - ΔV/Δr

Therefore , ΔV = - EΔr

Va-Vc = -4E or it can be written as Vc-Va = 4E

AC^{2} = AB^{2} + BC^{2}

AB^{2} = AC^{2} - BC^{2}

AB^{2} = 5^{2} - 3^{2}

AB = Δr = 4

(b) Vc-Va = 4E as its positivity shows Vc is greater than Va. So, it clearly shows electric potential is greater at point c because potential goes low along the direction of electric field.