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

`a=1.16* 10^(20)\ m/s^2`

Step-by-step explanation:

Given that,

First charge,
`q_1=50\ \mu C = 50* 10^(-6)\ C`. It is placed at y = 3 cm

Second charge,
`q_2=77\ \mu C = 77* 10^(-6)\ C` . It is placed at x = 4 cm.

Force on electron due to q₁.

`F_1=(kq_1e)/(r^2)\\\\F_1=(9* 10^9* 1.6* 10^(-19)* 50* 10^(-6))/((0.03)^2)\\\\F_1=8* 10^(-11)\ N`

Force on electron due to q₂.

`F_2=(kq_1e)/(r^2)\\\\F_2=(9* 10^9* 77* 10^(-6)* 1.6* 10^(-19))/((0.04)^2)\\\\F_2=6.93* 10^(-11)\ N`

The net force on the electron is given by :

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

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

F = ma

`a=(F)/(m)\\\\a=(10.58* 10^(-11))/(9.1* 10^(-31))\\\\a=1.16* 10^(20)\ m/s^2`

So, the acceleration of the electron is
`1.16* 10^(20)\ m/s^2`.

Answer 2

Step-by-step 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