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A student asked, “Since electrical potential is always proportional to potential energy, why bother with the concept of potential at all?” How would you respond?

Short Answer

Expert verified

The electric field is a conservative field and depending on the electrical potential- As We know to calculate the electric potential needs at least two charges one of them is a test charge. If We put a test charge in a medium of many charges, then the electric field generates around the test charge due to the surrounding charges.

Step by step solution

01

About electric potential 

The electric potential difference between points A and B, VB−VA is defined to be the change in potential energy of a charge q moved from A to B, divided by the charge.Units of potential difference are joules per coulomb, given the name volt (V) after Alessandro Volta.

02

Determine the potential

Solution:

As we know the electric field is a conservative field- Also, We know that the electric field is given by the following relation:

As we know in mechanics in general, We have two important concepts energy and potentiaL As We mention the electric field is a conservative field and depending on the electrical potentiaL As We know to calculate the electric potential needs at least two charges one of them is a test charge- If we put a test charge in a medium of many charges, then the electric filed generates around the test charge due to the surrounding charges- In general, We have electric potential and mechanical potential-

Therefore

The electric field is a conservative field and depending on the electrical potential- As We know to calculate the electric potential needs at least two charges one of them is a test charge. If We put a test charge in a medium of many charges, then the electric field generates around the test charge due to the surrounding charges.

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Most popular questions from this chapter

The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential difference. What is the resistance of (a) a 100-W bulb and (b) a 60-W bulb? (c) How much current does each bulb draw in normal use?

In the circuit of Fig. E25.30, the 5.0 Ω resistor is removed and replaced by a resistor of unknown resistance R. When this is done, an ideal voltmeter connected across the points band creads 1.9 V. Find (a) the current in the circuit and (b) the resistance R. (c) Graph the potential rises and drops in this circuit (see Fig. 25.20).

A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80

min. Silver containsfree electrons per cubic meter. (a) What is the

current in the wire? (b) What is the magnitude of thedrift velocity of the

electrons in the wire?

In the circuit shown in Fig. E26.20, the rate at which R1 is dissipating electrical energy is 15.0 W. (a) Find R1 and R2. (b) What is the emf of the battery? (c) Find the current through both R2 and the 10.0 Ω resistor. (d) Calculate the total electrical power consumption in all the resistors and the electrical power delivered by the battery. Show that your results are consistent with conservation of energy.

Copper has 8.5×1022free electrons per cubic meter. A 71.0-cm

length of 12-gauge copper wire that is 2.05 mm in diameter carries 4.85 A of

current. (a) How much time does it take for an electron to travel the length

of the wire? (b) Repeat part (a) for 6-gauge copper wire (diameter 4.12 mm)

of the same length that carries the same current. (c) Generally speaking,

how does changing the diameter of a wire that carries a given amount of

current affect the drift velocity of the electrons in the wire?

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