If the separation distance is doubled, then the electric field decreases by a factor of 4.
<h3>What is the electric field strength?</h3>
We know that the electric field strength is known to depend on the magnitude of the charge and the distance of separation. We know that the electric field refers to the region in which the influence of a charge is felt. Recall that a charge is a specie that is positively or negatively charged. The charge on a specie must always be shown by its sign.
We know that the electric field is the region in space where the influence of a charge can be felt. If a charge is placed in the vicinity of another charge, the second charge would experience a force due to the presence of the first charge. This is because the second charge was brought into the electric field of the first charge.
Thus we know that;
E = Kq/r^2
Where;
E = electric field strength
q = magnitude of charge
r = distance of separation
Now;
E = 9.0* 10^9 * 3.052 * 10^-6/(8.22 * 10^-2)^2
E = 4 N/C
Given that the electric filed strength is inversely proportional to the distance of separation, when the distance between the charges is doubled, the electric field decreases by a factor of 4.
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Explanation:
It is given that,
Mass of an electron, 
Initial speed of the electron, 
Final speed of the electron, 
Distance, d = 5 cm = 0.05 m
(a) The acceleration of the electron is calculated using the third equation of motion as :
Force exerted on the electron is given by :
(b) Let W is the weight of the electron. It can be calculated as :
Comparison,
Hence, this is the required solution.
Answer:
Explanation:
Given
Pressure, Temperature, Volume of gases is
Let P & T be the final Pressure and Temperature
as it is rigid adiabatic container therefore Q=0 as heat loss by one gas is equal to heat gain by another gas
where Q=heat loss or gain (- heat loss,+heat gain)
W=work done by gas
change in internal Energy of gas
Thus from 1 & 2 we can say that
where 
and 
Answer:
The force of the nail pushing in the opposite direction
Answer:
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