500J work is required to carry charged particle between two points with potential difference of 20V. what is the magnitude of charge on particle?
Answers
Answer: According to Question
Data
Work required ΔU = 500J
Potential difference ΔV =20 V
Magnitude of Charge ?
Explanation:
By using formula
ΔV = ΔU/Q
or Q = ΔU/ΔV
by putting values
Q = 500/20
Q = 25 Coloumb
Answer:
The magnitude of the charge on the particle is 25 Coulomb.
Explanation:
The change in potential energy is the work done. The potential energy at a point is defined as the work done to bring a charge from infinity to the observable point.
Potential energy is not defined rather change in potential energy is defined.
and Work done = change in potential energy = charge X potential
i.e. W = q X V
GIVEN
work done = 500 J=W
potential = 20 v =V
charge =q =?
So, putting the value in the formula, we will get:
⇒500 = q X 20
⇒q = 25 C
Therefore, the magnitude of the charge is 25 coulomb.