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GREYUIT [131]
3 years ago
10

Four particles are positioned on the rim of a circle. The charges on the particles are 10.500 mC, 11.50 mC, 21.00 mC, and 20.500

mC. If the electric potential at the center of the circle due to the 10.500 mC charge alone is 4.50 3 104 V, what is the total electric potential
Physics
1 answer:
gregori [183]3 years ago
7 0

The total electric potential will be 4.50 ×10⁴ V. The potential difference is the difference in electric potential between two charged substances.

<h3>What is the electrical potential difference?</h3>

The amount of work performed in an electrical field to move a unit charge from one location to another is defined as the electrical potential difference.

The distance between all charges and the center is the same. The 10,500 mC, 11.50 mC, 21.00 mC, and 20,500 mC charges neglect each other's potentials.

Hence, the total electric potential will be 4.50 ×10⁴ V.

To learn more about the electric potential difference, refer to the link;

brainly.com/question/9383604

#SPJ1



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A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)
bezimeni [28]

Answer:

205 V

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Explanation:

L = Inductance in Henries, (H)  = 0.500 H

resistor is of 93 Ω so R = 93 Ω

The voltage across the inductor is

V_{L} = - IwLsin(wt)

w = 500 rad/s

IwL = 11.0 V

Current:

I = 11.0 V / wL

 = 11.0 V / 500 rad/s (0.500 H)

 = 11.0 / 250

I = 0.044 A

Now

V_{R} = IR

    = (0.044 A) (93 Ω)

V_{R} = 4.092 V

Deriving formula for voltage across the resistor

The derivative of sin is cos

V_{R} = V_{R} cos (wt)

Putting V_{R} = 4.092 V and w = 500 rad/s

V_{R} = V_{R} cos (wt)

    = (4.092 V) (cos(500 rad/s )t)

So the voltage across the resistor at 2.09 x 10-3 s is which means

t = 2.09 x 10⁻³

V_{R} = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))

    =  (4.092 V) (cos (500 rads/s)(0.00209))

    = (4.092 V) (cos(1.045))

    = (4.092 V)(0.501902)

    = 2.053783

V_{R} = 2.05 V

8 0
3 years ago
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