Electrostatics

Question: The radius of nucleus of silver (atomic number = 47) is $3.4 \times {10^{ - 14}}m$. The electric potential on the surface of nucleus is $(e = 1.6 \times {10^{ - 19}}C)$



1) $1.99 \times {10^6}\,volt$
2) $2.9 \times {10^6}\,volt$
3) $4.99 \times {10^6}\,volt$
4) $0.99 \times {10^6}\,volt$
Solution: Explanation: Explanation
electric-field-potential

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