Question

A charge of 50-μC is placed on the y axis at y = 3.0 cm and a 77-μC charge is placed on the x axis at x = 4.0 cm. If both charges are held fixed, what is the magnitude of the initial acceleration of an electron (me = 9.1 x 10-31 kg) released from rest at the origin?

Answer 1

Answer:

Explanation:

We shall first calculate the electric field at origin .

Field due to first charge

= KQ / R²

= 9  x 10⁹ x 50 x 10⁻⁶ / .03²

E₁ = 5 x 10⁸ N/C

Field due to second charge

= KQ / R²

= 9  x 10⁹ x 77 x 10⁻⁶ / .04²

E₂ = 4.33  x 10⁸ N/C

Resultant field E = √ ( 5² + 4.33² )  x 10⁸

= 6.6 x 10⁸ N/C

Force on electron = E x q

= 6.6 x 10⁸ x 1.6 x 10⁻¹⁹ N

= 10.56 x 10⁻¹¹ N

Acceleration = force / mass

=   10.56 x 10⁻¹¹ / 9.1 x 10⁻³¹

= 1.16 x 10²⁰ m /s²

Answer 2

Answer:

$a=1.16\times 10^{20}\ m/s^2$

Explanation:

Given that,

First charge, $q_1=50\ \mu C = 50\times 10^{-6}\ C$. It is placed at y = 3 cm

Second charge, $q_2=77\ \mu C = 77\times 10^{-6}\ C$ . It is placed at x = 4 cm.

Force on electron due to q₁.

$F_1=\dfrac{kq_1e}{r^2}\\\\F_1=\dfrac{9\times 10^9\times 1.6\times 10^{-19}\times 50\times 10^{-6}}{(0.03)^2}\\\\F_1=8\times 10^{-11}\ N$

Force on electron due to q₂.

$F_2=\dfrac{kq_1e}{r^2}\\\\F_2=\dfrac{9\times 10^9\times 77\times 10^{-6}\times 1.6\times 10^{-19}}{(0.04)^2}\\\\F_2=6.93\times 10^{-11}\ N$

The net force on the electron is given by :

$F=\sqrt{F_1^2+F_2^2} \\\\F=\sqrt{(8\times 10^{11})^2+(6.93\times 10^{-11})^2} \\\\F=1.058\times 10^{-11}\ N\\\\\text{or}\\\\F=10.58\times 10^{-11}\ N$

Let a is the acceleration of the electron. Net force is given by :

F = ma

$a=\dfrac{F}{m}\\\\a=\dfrac{10.58\times 10^{-11}}{9.1\times 10^{-31}}\\\\a=1.16\times 10^{20}\ m/s^2$

So, the acceleration of the electron is $1.16\times 10^{20}\ m/s^2$.