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Olegator
2 months ago
6

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 pote

ntial 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
Physics
2 answers:
Keith_Richards [3.2K]2 months ago
8 0
1) The electric potential energy can be defined as the product of the electric potential and the associated charge:
U=q V
where
q refers to the charge
V denotes the electric potential

In this scenario, the charge on the rod is q=4.0 \mu C = 4.0 \cdot 10^{-6}C, and the potential energy is U=5.2 \cdot 10^{-3} J, thus we may rearrange the earlier formula to find 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) Using this same formula, if the charge changes to q=2.0 \mu C = 2.0 \cdot 10^{-6} C, the resulting 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
Ostrovityanka [3.2K]2 months ago
3 0

Clarification:

Considering 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 relationship between electric potential and electric potential energy is established by:

U=qV

CASE 1

Thus,

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

In instances where the charge is 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

Consequently, this presents the necessary solution.

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