Electric potential at the conductor of ball – problems and solutions

1. A 4-μC hollow ball conductor has radius of 8-cm. Determine the electric potential at the surface of the ball. (k = 9.109 N.m2.C-2)

Known :

The electric charge (Q) = 4 μC = 4 x 10-6 C Electric potential at the conductor of ball – problems and solutions 1

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

Coulomb’s constant (k) = 9.109 N.m2.C-2

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

Solution :

V = k Q / r

V = (9 x 109)(4 x 10-6) / (8 x 10-2)

V = (36 x 103) / (8 x 10-2)

V = (36/8) x 103 x 102

V = 4.5 x 105 Volt

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2. A spherical conductor has radius of 3-cm (1 μC = 10-6 C and k = 9.109 N.m2.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 Electric potential at the conductor of ball – problems and solutions 2

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

Coulomb’s constant (k) = 9.109 N.m2.C-2

Wanted : The electric potential at point A (V)

Solution :

V = k Q / r

V = (9 x 109)(1 x 10-6) / (3 x 10-2)

V = (9 x 103) / (3 x 10-2)

V = (9/3) x 103 x 102

V = 3 x 105 Volt

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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 = 9.109 N.m2.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 CElectric potential at the conductor of ball – problems and solutions 3

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

Coulomb’s constant (k) = 9.109 N.m2.C-2

Wanted : The electric potential at point P (V)

Solution :

V = k Q / r

V = (9 x 109)(6,4 x 10-9) / (9 x 10-2)

V = 109 (6,4 x 10-9) / 10-2

V = 6.4 / 10-2

V = 6.4 x 102

V = 6.4 x 100

V = 640 Volt

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