Question

Electric potential varies with distance such that V(x) =ax-bx^3; where a and b are constants. Where will the electric field intensity be zero?

(a) x=\(\frac {a}{b}\)

(b) x=\(\frac {a}{3b}\)

(c) x=\(\sqrt{\frac {a}{b}}\)

(d) x=+\(\sqrt{\frac {a}{3b}}\) and x=-\(\sqrt{\frac {a}{3b}}\)

I got this question in semester exam.

The query is from Electrostatic Potential in division Electrostatic Potential and Capacitance of Physics – Class 12

Answer 1

The correct option is (d) x=+\(\sqrt{\frac {a}{3b}}\) and x=-\(\sqrt{\frac {a}{3b}}\)

To explain: We know that E=-\(\frac {dv}{dx}\)

∴ E = –\(\frac {d}{dx}\) (ax – bx^3) = -(a – 3bx^2)

Electric field will be zero. ∴ E = 0 ⇒ (a-3bx^2) = 0

⇒ a = 3bx^2

⇒ x^2 = \(\frac {a}{3b}\) = x = ±\(\sqrt{\frac {a}{3b}}\)