First slide

Electric field

Question

Three point charges + q each are fixed at $A (L / \sqrt{2}, L / \sqrt{2}, 0), B (- L / \sqrt{2}, L / \sqrt{2}, 0), C (0, - L, 0)$ . A fourth point charge – q and mass ‘m’ is kept on z-axes at D (0,0,Z) with Z<<L and released, the minimum time taken by the fourth charge to cross the xy-plane is nearly (neglect effect of gravity)

Difficult

A

$2 \sqrt{\frac{π^{3} \in_{0} {mL}^{3}}{3 q^{m} 2 m}}$
B

$\sqrt{\frac{π^{3} \in_{0} {mL}^{3}}{q^{2 m}}}$
C

$\sqrt{\frac{2 π^{3} \in_{0} m L^{3}}{3 q^{2}}}$
D

$\sqrt{\frac{π^{3} \in_{0} m L^{3}}{3 q^{2}}}$

Solution

$F_{net} = 3 F \cos θ$
$= - 3 \frac{1}{4 π ϵ_{0}} \frac{q^{2}}{r^{2}} (\frac{z}{r})$
$= \frac{- 3 q^{2} z}{4 π \in_{0} r^{3}} \approx \frac{- 3 q^{2} z}{4 π \in_{0} L^{3}} \{\begin{matrix} ∵ r = \sqrt{L^{2} + z^{2}} \\ r \approx L for z << L \end{matrix}\}$
$\begin{array}{l} ma = - \frac{3 q^{2} z}{4 π \in_{0} L^{3}} \\ a = - ω^{2} z \\ T = \frac{2 π}{ω} = 2 π \sqrt{\frac{4 π ϵ_{0} {mL}^{3}}{3 q^{2}}} \\ t = \frac{T}{4} = \sqrt{\frac{π^{3} \in_{0} {mL}^{3}}{3 q^{2}}} \end{array}$

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