Question

A hard rubber rod with an electric potential energy of 5.2 × 10–3 J has a charge of 4.0 µC at the tip. What is the electric potential at the tip? Round your answer to one decimal place. × 103 V What is the electric potential if the charge at the tip changes to 2.0 µC? Round your answer to one decimal place. × 103 V

Answer 1

1) The electric potential energy is equal to the product between the electric potential and the charge:
$U=q V$
where
q is the charge
V is the electric potential

In our problem, the charge on the rod is $q=4.0 \mu C = 4.0 \cdot 10^{-6}C$, while its potential energy is $U=5.2 \cdot 10^{-3} J$, therefore we can re-arrange the previous formula to get the electric potential at the tip:
$V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3}J}{4.0 \cdot 10^{-6} C}=1300 V=1.3 \cdot 10^3 V$

2) By using the same formula, If the charge is changed to $q=2.0 \mu C = 2.0 \cdot 10^{-6} C$, the electric potential will be:
$V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3} J}{2.0 \cdot 10^{-6}C}=2600 V = 2.6 \cdot 10^{3}V$

Answer 2

Explanation :

Given that,

Electric potential energy, $U=5.2\times 10^{-3}\ J$

Charge on a rubber rod, $q=4\ \mu C=4\times 10^{-6}\ C$

The relation between the electric potential and electric potential energy is given by :

$U=qV$

CASE 1

So,

$V=\dfrac{U}{q}$

$V=\dfrac{5.2\times 10^{-3}\ J}{4\times 10^{-6}\ C}$

$V=1300\ V$

or

$V=1.3\times 10^3\ V$

CASE 2

If charge, $q=2\ \mu C=2\times 10^{-6}\ C$

$V=\dfrac{U}{q}$

$V=\dfrac{5.2\times 10^{-3}\ J}{2\times 10^{-6}\ C}$

$V=2600\ V$

or

$V=2.6\times 10^3\ V$

Hence, this is the required solution.