Question

A proton is moving at 105 m/s at a point where the potential is 10 V. Later, it is at a place where the potential is 5 V. What is its speed there, assuming energy is conserved?

Answer 1

Answer:

The speed is $7.07\times10^{4}\ m/s$

Explanation:

Given that,

Speed of proton $v= 10^{5}\ m/s$

Final potential = 10 v

Initial potential = 5 V

We need to calculate the speed

Using formula of energy

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

$v^2=\dfrac{2eV}{m}$

The speed of the particle is directly proportional to the potential.

$v^2\propto V$

Put the value into the formula

$(10^{5})\propto 10$....(I)

For 5 V,

$v^2\propto 5$.....(II)

From equation (I) and (II)

$\dfrac{(10^{5})^2}{v^2}=\dfrac{10}{5}$

$v=70710.67\ m/s$

$v=7.07\times10^{4}\ m/s$

Hence, The speed is $7.07\times10^{4}\ m/s$