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 released from rest at the origin, in units of 1020 m/s2?

Answer 1

Answer:

The acceleration of an electron is $1.2\times10^{20}\ m/s^2$

Explanation:

Given that,

One Charge = 50 μC

Distance on y axis = 3.0 cm

Second charge = 77 μC

Distance on x axis = 4.0 cm

We need to calculate the force on electron due to q₁

Using formula of force

$F_{1}=\dfrac{kq_{1}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}$

$F_{1}=8\times10^{-11}j\ \ N$

We need to calculate the force on electron due to q₂

Using formula of force

$F_{2}=\dfrac{kq_{2}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}$

$F_{2}=6.93\times10^{-11}i\ \ N$

We need to calculate the net force

Using formula of net force

$F=F_{1}+F_{2}$

Put the value into the formula

$F=8\times10^{-11}j+6.93\times10^{-11}i$

The magnitude of the net force

$F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}$

$F=1.058\times10^{-10}\ N$

We need to calculate the acceleration of an electron

Using formula of force

$F = ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}$

$a=1.2\times10^{20}\ m/s^2$

Hence, The acceleration of an electron is $1.2\times10^{20}\ m/s^2$