Question

A stationary proton is moved form point A, where the potential is 459.0 V, to point B, where the potential is 125.0 V. (a) How much work is done be the electric force? (b) What is its speed to point B

Answer 1

TWO PROTONS A AND B ARE PLACED BETWEEN TWO PARALLEL PLATES HAVING A POTENTIAL DIFFERENCE V AS SHOWN IN THE FIGURE.

WHAT IS THE RATIO OF ELECTRIC FIELD INTENSITIES AT POINTS A AND B BETWEEN THE PLATES OF A PARALLEL PLATE CAPACITOR?

WILL THESE PROTON EXPERIENCE EQUAL OR UNEQUAL FORCE? JUSTIFY?

Answer 2

The work done by the electric force, $W=5.34\times 10^{-17}\ J$

The speed of the proton when it reaches to point B is $1.54\times 10^5\ m/s$

Explanation :

It is given that,

Potential at point A, $V_A=459\ V$

Potential at point B, $V_B=125\ V$

(a) We need to find the work done by the electric force. It is equal to the product of charge and potential difference such that,

$W=q(V_B-V_A)$

$W=1.6\times 10^{-19}\times (459-125)$

$W=5.34\times 10^{-17}\ J$

(b) Let v is the speed of the proton with which it reaches at point B. It can be calculated using conservation of energy as :

$\dfrac{1}{2}mv^2=qV$

$v=\sqrt{\dfrac{2qV}{m}}$

m is the mass of proton

V is the potential at B

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 125}{1.67\times 10^{-27}}}$

$v=1.54\times 10^5\ m/s$

So, the speed of the proton when it reaches to point B is $1.54\times 10^5\ m/s$. Hence, this is the required solution.

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Electric potential