A total charge of 5 microcoulomb is distributed uniformly on the surface of the thin walled hemispherical Cup. If the electric field strength at the centre of the hemisphere is 9 into 10 raise to power 8 Newton per coulomb then the radius of the cup is A.5mm B.10mm C.5cm D.10cm?

Question

Answer

Solution:

Given:

Total charge (Q) = 5 μC = 5 × 10^-6 C

Electric field strength (E) = 9 × 10⁸ N/C

Concept:

The electric field at the center of a uniformly charged hemispherical shell is given by:

E = Q/(4πε₀R²)

Where Q is the total charge on the shell, R is the radius of the shell, and ε₀ is the permittivity of free space

Calculation:

Let the radius of the hemispherical cup be R

Total charge on the surface of the cup (Q) = 5 × 10^-6 C

Electric field strength at the center of the cup (E) = 9 × 10⁸ N/C

Permittivity of free space (ε₀) = 8.85 × 10^-12 C²/N m²

Using the formula for electric field, we have:

E = Q/(4πε₀R²)

R² = Q/(4πε₀E)

R² = (5 × 10^-6)/(4π × 8.85 × 10^-12 × 9 × 10⁸)

R² = 0.000159

R = 0.0126 m = 12.6 mm

Answer:

The radius of the hemispherical cup is 12.6 mm (approx)

Answer

I got radius^2=5×10^-5metre

Answer

Option B 10 mm

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