Question

An electron is released at rest from point A. It moves in response to the electric field of a fixed charge distribution along a path that takes it through point B. If the potential has a value of -11 V at point A and 25 V at point B, what is the electron's speed as it passes point B?

Answer 1

Answer:

$v=8.89\times 10^{15}\ m/s$

Explanation:

An electron is released at rest from point A. It moves in response to the electric field of a fixed charge distribution along a path that takes it through point B.

Potential at point A, $V_A=-11\ V$

Potential at point B, $V_B=25\ V$

Let v is the speed of the electron as it passes point B. It can be calculated using the conservation of energy theorem as :

$\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=V_B-V_A$

Here, u = 0

$\dfrac{1}{2}mv^2=25-(-11)$

$mv^2=72$

$v=\sqrt{\dfrac{72}{m}}$

m is the mass of electron

$v=\sqrt{\dfrac{72}{9.1\times 10^{-31}}}$

$v=8.89\times 10^{15}\ m/s$

So, the electron's speed as it passes point B is $8.89\times 10^{15}\ m/s$. Hence, this is the required solution.