Electric field intensity and electric potential at a certain distance from a point charge is 32 N/C and 16 J/C. What is the distance from the charge?a)50 mb)0.5 mc)10 md)7 mCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

Given:
Electric field intensity (E) = 32 N/C
Electric potential (V) = 16 J/C

To find: Distance from the point charge

Formulae:
Electric field intensity (E) = 1/4πε₀ * Q/r²
Electric potential (V) = 1/4πε₀ * Q/r

Where,
Q = Point charge
r = Distance from the point charge
ε₀ = Permittivity of free space

Solution:
Let's find Q first using the given values of E and V.

From the formula for electric field intensity (E), we have
E = 1/4πε₀ * Q/r²
So, Q = E * 4πε₀ * r²

Substituting the given values, we get
Q = 32 * 4πε₀ * r²

Now, using the formula for electric potential (V), we have
V = 1/4πε₀ * Q/r
Substituting the value of Q, we get
V = 1/4πε₀ * 32 * 4πε₀ * r² / r
V = 8 * r

Substituting the given value of V, we get
16 = 8 * r
r = 2 m

Therefore, the distance from the point charge is 0.5 m (Option B).

Answer

Electric field due to a charge q at a distance r is

and electric potential at that point is

Therefore,

and

Dividing these two equations, we get r=0.5m. If the medium is air, k=1. Thus we can get the value of q=0.89 C.

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