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

Explanation :

Given that,

Electric potential energy,
`U=5.2* 10^(-3)\ J`

Charge on a rubber rod,
`q=4\ \mu C=4* 10^(-6)\ C`

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

`U=qV`

CASE 1

So,

`V=(U)/(q)`

`V=(5.2* 10^(-3)\ J)/(4* 10^(-6)\ C)`

`V=1300\ V`

or

`V=1.3* 10^3\ V`

CASE 2

If charge,
`q=2\ \mu C=2* 10^(-6)\ C`

`V=(U)/(q)`

`V=(5.2* 10^(-3)\ J)/(2* 10^(-6)\ C)`

`V=2600\ V`

or

`V=2.6* 10^3\ V`

Hence, this is the required solution.

Answer 2

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= (U)/(q)= (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= (U)/(q)= (5.2 \cdot 10^(-3) J)/(2.0 \cdot 10^(-6)C)=2600 V = 2.6 \cdot 10^(3)V`