1. A 4-μC hollow ball conductor has radius of 8-cm. Determine the electric potential at the surface of the ball. (k = 9.10^{9} N.m^{2}.C^{-2})

__Known :__

The electric charge (Q) = 4 μC = 4 x 10^{-6} C

The radius of ball (r) = 8 cm = 8 x 10^{-2} m

Coulomb’s constant (k) = 9.10^{9} N.m^{2}.C^{-2}

__Wanted :__ The electric potential at the surface of the ball (V)

__Solution :__

V = k Q / r

V = (9 x 10^{9})(4 x 10^{-6}) / (8 x 10^{-2})

V = (36 x 10^{3}) / (8 x 10^{-2})

V = (36/8) x 10^{3} x 10^{2}

V = 4.5 x 10^{5 }Volt

[irp]

2. A spherical conductor has radius of 3-cm (1 μC = 10^{-6 }C and k = 9.10^{9} N.m^{2}.C^{-2}). If the conductor is positively charged +1 μC then the electric potential at point A is …

Solution :

*The electric potential inside the spherical conductor = The electric potential at the surface of the spherical conductor.*

__Known :__

The electric charge (Q) = 1 μC = 1 x 10^{-6} C

The radius of the spherical conductor (r) = 3 cm = 3 x 10^{-2} m

Coulomb’s constant (k) = 9.10^{9} N.m^{2}.C^{-2}

__Wanted :__ The electric potential at point A (V)

__Solution :__

V = k Q / r

V = (9 x 10^{9})(1 x 10^{-6}) / (3 x 10^{-2})

V = (9 x 10^{3}) / (3 x 10^{-2})

V = (9/3) x 10^{3} x 10^{2}

V = 3 x 10^{5 }Volt

[irp]

3. A hollow metal ball with radius of 9-cm has 6.4 x 10^{-9} Coulomb electric charge, as shown in figure below. Distance between point O and point P = 4 cm; Distance between point P and point Q = 5 cm; Distance between point Q and point R = 18 cm and k = 9.10^{9} N.m^{2}.C^{-2}. Determine the electric potential at point P.

Solution

*The electric potential inside the spherical conductor = The electric potential at the surface of the spherical conductor.*

__Known :__

The electric charge (Q) = 6.4 x 10^{-9} C

*The radius of the spherical conductor* (r) = OP + PQ = 4 cm + 5 cm = 9 cm = 9 x 10^{-2 }m

Coulomb’s constant (k) = 9.10^{9} N.m^{2}.C^{-2}

__Wanted :__ The electric potential at point P (V)

__Solution :__

V = k Q / r

V = (9 x 10^{9})(6,4 x 10^{-9}) / (9 x 10^{-2})

V = 10^{9 }(6,4 x 10^{-9}) / 10^{-2}

V = 6.4 / 10^{-2}

V = 6.4 x 10^{2}

V = 6.4 x 100

V = 640 Volt

[irp]