A conducting sphere of radius R-20cm is given a charge Q-16- C- What is E at centre

The correct option is A 3.6×10^6N/C R = 20 cm = 20 × 10^ 2m Q = 16 μ C = 16 × 10^ 6C E = 14πε_0 . QR^2 = 9 × 10^9× 16 × 10^ 620 × 20 × 10^ 4 ...

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Standard XII

Physics

Surface Integral

A conducting ...

Question

A conducting sphere of radius $R = 20 c m$ is given a charge $Q = 16 μ C$ . What is $\to E$ at centre

3.6 \times 10^{6} N / C

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1.8 \times 10^{6} N / C

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zero

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0.9 \times 10^{6} N / C

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Solution

The correct option is A $3.6 \times 10^{6} N / C$
$R = 20 c m = 20 \times 10^{- 2} m$
$Q = 16 μ C = 16 \times 10^{- 6} C$

$E = \frac{1}{4 π ε_{0}} . \frac{Q}{R^{2}} = \frac{9 \times 10^{9} \times 16 \times 10^{- 6}}{20 \times 20 \times 10^{- 4}} = 3.6 \times 10^{6} μ / C$

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A conducting sphere of radius R=20cm is given a charge Q=16μC. What is →E at centre

The correct option is A 3.6×106N/CR=20cm=20×10−2mQ=16μC=16×10−6CE=14πε0.QR2=9×109×16×10−620×20×10−4=3.6×106μ/C

A conducting sphere of radius $R = 20 c m$ is given a charge $Q = 16 μ C$ . What is $\to E$ at centre

The correct option is A $3.6 \times 10^{6} N / C$
$R = 20 c m = 20 \times 10^{- 2} m$
$Q = 16 μ C = 16 \times 10^{- 6} C$

$E = \frac{1}{4 π ε_{0}} . \frac{Q}{R^{2}} = \frac{9 \times 10^{9} \times 16 \times 10^{- 6}}{20 \times 20 \times 10^{- 4}} = 3.6 \times 10^{6} μ / C$