Get a direct answer of what???
Answer:
Work = 5941 J
Explanation:
As we know that work done is given by the equation

here we know that

also we have

now from above formula we have


Answer:
hey mate
answer is probably voltage as per me
as
Explanation:
Voltage, electric potential difference, electric pressure or electric tension is the difference in electric potential between two points, which is defined as the work needed per unit of charge to move a test charge between the two points
Answer:
The resistance that will provide this potential drop is 388.89 ohms.
Explanation:
Given;
Voltage source, E = 12 V
Voltage rating of the lamp, V = 5 V
Current through the lamp, I = 18 mA
Extra voltage or potential drop, IR = E- V
IR = 12 V - 5 V = 7 V
The resistance that will provide this potential drop (7 V) is calculated as follows:
IR = V

Therefore, the resistance that will provide this potential drop is 388.89 ohms.
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q = 
=
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is
.