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

Electric potential varies with distance such that V(x) =ax-bx^3; where

1.	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» 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 = ±\(\sqrt{\frac {a}{3b}}\)

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