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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

Respuesta :

skyluke89
1) The electric potential energy is equal to the product between the electric potential and the charge:
[tex]U=q V[/tex]
where
q is the charge
V is the electric potential

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

2) By using the same formula, If the charge is changed to [tex]q=2.0 \mu C = 2.0 \cdot 10^{-6} C[/tex], the electric potential will be:
[tex]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 [/tex]

Explanation :

Given that,

Electric potential energy, [tex]U=5.2\times 10^{-3}\ J[/tex]

Charge on a rubber rod, [tex]q=4\ \mu C=4\times 10^{-6}\ C[/tex]

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

[tex]U=qV[/tex]

CASE 1

So,

[tex]V=\dfrac{U}{q}[/tex]

[tex]V=\dfrac{5.2\times 10^{-3}\ J}{4\times 10^{-6}\ C}[/tex]

[tex]V=1300\ V[/tex]

or

[tex]V=1.3\times 10^3\ V[/tex]

CASE 2

If charge, [tex]q=2\ \mu C=2\times 10^{-6}\ C[/tex]

[tex]V=\dfrac{U}{q}[/tex]

[tex]V=\dfrac{5.2\times 10^{-3}\ J}{2\times 10^{-6}\ C}[/tex]

[tex]V=2600\ V[/tex]

or

[tex]V=2.6\times 10^3\ V[/tex]

Hence, this is the required solution.

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