An electron is placed midway between two fixed charges, q1 = 2.5 × 10-10 C and q2 = 5 × 10-10 C. If the charges are 1 m apart, what is the velocity of the electron when it reaches a point 10cm from q2?

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Solution: Here, given two charges areUser Loaded Image | CSIT Guide

q1 = 2.5 × 10-10 C

q2  = 5 × 10-10 C

Distance between  qand q2

r = 1 m

Distance between e and q1,q2 = r1 = r2 = 0.5 m

Distance travelled by e towards q2 from the initial position

S = 0.5 m – 0.1 m

= 0.4 m

Now we know that

From equation of motion

v2 = u2 + 2as                          Since, u = 0 ms-1

\( v = \sqrt{2as}\)   ………………… (1)

Again, Electrostatic force between qand e at rest condition

F1 = \( \frac{q_{1}e}{4\pi \epsilon_{o} r_{1}^{2}}\)

Electrostatic force between q2 and e

F2 = \( \frac{q_{2}e}{4\pi \epsilon_{o} r_{1}^{2}}\)

Now, Net force F = F2 – F1 = \(\frac{q_{1}e}{4\pi \epsilon_{o}r_{1}^{2}} – \frac{q_{2}e}{4\pi \epsilon_{o}r_{2}^{2}}\)

F = \( \frac{e}{4\pi \epsilon_{o} }\) = \([\frac{q_2}{r_{2}^{2}} – \frac{q_1}{r_{1}^{2}}]\) = \(\frac{1.6 × 10^{-19} × 9 × 10^{9}}{\left( 0.5 \right)^{2}} × [ 5 × 10^{-10} – 2.5 × 10^{-10}]\)

= 1.44 × 10-18 N

From Newton’s second law of motion

F = ma = 1.44 × 10-18

or, a = \(\frac{1.44 × 10^{-19}}{9.1 × 10^{-31}}\)              [∴ m = me = 9.1 × 10-31 kg]

∴ a = 1.58 × 1012 ms-2

Substituting value of ‘a’ in equation (1), we get

v = \(\sqrt{2 × 1.58 × 10^{12} × 0.4}\)

∴ v = 1.125 × 106 ms-1

Hence, required velocity of electron when it reaches a point 10 cm from q2 is 1.125 × 106 ms-1.

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