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3 years ago

What is the difference in electrical potential energy between two places in an electric field?

Physics

2 answers:

Phoenix [80]3 years ago

4 0

Answer: Option (c) is the correct answer.

Explanation:

The difference in electrical potential energy between two places in an electric field is known as potential difference.

Mathematically, potential difference between two given points can be written as follows.

$\Delta V = V_{2} - V_{1}$

where, $\Delta V$ = potential difference

$V_{1}$ = potential at point 1

$V_{1}$ = potential at point 2

ASHA 777 [7]3 years ago

4 0

Option (c) is correct. The difference between the potential energy of the two points in an electric field is called as potential difference.

Further Explanation:

The electric potential energy is the amount of energy required to make a charge move against the electric field present in the region.

The difference between the measures of the potential energy between the two points present in an electric field is termed as the potential difference between those two points as it measures the difference in energy required for the movement of charge particle from one point to the other.

The mathematical expression of the potential difference shows the above statement to be true.

$\Delta V = {V_2} - {V_1}$

Here, $\Delta V$ is the potential difference between the two points, $V_1$ is the potential of first point and $V_2$ is the potential of second point.

Thus, the difference in the potential energy of the two points in the electric field shows the potential difference between the two points.

Learn More:

Simplify the circuit of three resistors and inductor as shown brainly.com/question/1695461
The frequency of the signals given by the FM radio station is brainly.com/question/9527365
The electrical shock is experienced by the body when brainly.com/question/3059888

Answer Details:

Grade: High School

Chapter: Electric current

Subject: Physics

Keywords: Electric potential energy, difference, work done, moving, unit charge, potential difference, between two points, electric field, against.

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Read 2 more answers

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The given parameters;

first charge, q₁ = 8.75 μC
second charge, q₂ = -6.5 μC
electric field, E = 1.85 x 10⁸ N/C
distance between the two charges, r = 2.5 cm

(a)

The attractive force between the charges is calculated as follows;

$F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N$

The force on the negative charge due to the electric field is calculated as follows;

$F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N$

The tension on the wire is the resultant of the two forces and it is calculated as follows;

$T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N$

(b) when the two charges are negative

The repulsive force between the two charges is calculated as follows;

$F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N$

The force on the first negative charge due to the electric field is calculated as follows;

$F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N$

The force on the second negative charge due to the electric field is calculated as follows;

$F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N$

The tension on the wire is the resultant of the three forces and it is calculated as follows;

$T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N$

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3 years ago