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.
1 answer:
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 :
(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :
Therefore, this is the required solution.
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Answer:
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Explanation:
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Answer: B
Explanation: look at the chart, easy
Answer:
E= 4.35*10^6 N/C
Explanation:
Let's find the area charge density of the plate
α= 6.9*10^-6/9*10^-2 = 7.7*10^-5C/m2
Now we can calculate the electric field just of the plate
E =α/2e =7.7*10^-5/2*8.85*10^-12 = 4.35*10^6 N/C