Location C is 0.02 m from a small sphere which has a charge of 3 nanocoulombs uniformly distributed on its surface. Location D i
2 answers:
The change in potential along a path from C to D due to a small charged sphere is 900 V.
Given:
Charge, Q = 3 nC = 3 × 10⁻⁹ C
Distance between the sphere and point C, r₁ = 0.02 m
Distance between the sphere and point D, r₂ = 0.06 m
Calculation:
We know that the electric potential is given as:
V = k Q/r - (1)
where, V is the electric potential
k is Coulomb's force constant
Qis the charge on the sphere
r is the separation distance
The electric potential at point C due to charged sphere can be given as:
V₁ = k Q/r₁
= (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.02 m)]
= 1350 V
The electric potential at point D due to charged sphere can be given as:
V₂ = k Q/r₂
= (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.06 m)]
= 450 V
Now, the change in potential along the path from C to D can be calculated as:
ΔV = V₂ - V₁
= 450 V - 1350 V
= -900 V
The negative sign indicates that the work is done against the electric field in moving the charge from C to D.
Therefore, the change in potential along a path from C to D is 900 V against the direction of the electric field.
Learn more about the electric potential here:
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Hi there!
We can use the equation for the charge of a charging capacitor:
Using Capacitor equations:
Therefore, Cε equals the steady-state charge of the capacitor (the function approaches this value as t ⇒ ∞.
We can plug in the givens and solve.