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cluponka [151]
3 years ago
5

Find the voltage change when: a. An electric field does 12 J of work on a 0.0001-C charge. b. The same electric field does 24 J

of work on a 0.0002-C charge.
Physics
1 answer:
kondaur [170]3 years ago
7 0

Explanation:

Given that,

(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :

V=\dfrac{W}{q}

V=\dfrac{12}{0.0001}

V=12\times 10^4\ Volt

(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :

V=\dfrac{W}{q}

V=\dfrac{24}{0.0002}

V=12\times 10^4\ Volt

Therefore, this is the required solution.

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Magnification will be equal to 3

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We have given focal length of the converging lens F_1=20cm

Focal length of the diverging lens F_2=30cm

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Combination of the focal length will be equal to

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So combination of the focal length will be 60 cm

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<h3>What is wavelength?</h3>

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Based on the image, the following are wavelengths;

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