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3 years ago

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A proton is moved from a position where the electric potential is 125 V to a position where the electric potential is 275 V. The

magnitude of the charge on a proton is 1.602 × 10-19 C. What is the change in electric potential energy of the proton?
2.0 × 10-17 J
2.4 × 10-17 J
4.4 × 10-17 J
6.4 × 10-17 J

1 answer:

kakasveta [241]3 years ago

7 0

We determine the electric potential energy of the proton by multiplying the net electric potential to the charge of the proton. The net electric potential is the difference of the final state to the that of the initial state. So, it would be 275 - 125 = 150 V.

electric potential energy = 150 (<span>1.602 × 10-19) = 2.4x10^-17 J</span>

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A 20 cm-radius ball is uniformly charged to 71 nC.

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Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

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