A conducting sphere of 3.0 cm radius is charged with – 40 nC. In what follows, the electric potential is taken to be zero at infinity. a) Calculate the electric potential of the sphere. (2 points) b) Calculate the electric potential at 6.0 cm from the center of the sphere. (2 points)
The Correct Answer and Explanation is:
Solution
Given:
Radius r1=3.0 cm=0.03 mr_1 = 3.0 \, \text{cm} = 0.03 \, \text{m}r1=3.0cm=0.03m
Charge Q=−40 nC=−40×10−9 CQ = -40 \, \text{nC} = -40 \times 10^{-9} \, \text{C}Q=−40nC=−40×10−9C
k=14πε0=9.0×109 N\cdotpm2/C2k = \dfrac{1}{4 \pi \varepsilon_0} = 9.0 \times 10^9 \, \text{N·m}^2/\text{C}^2k=4πε01=9.0×109N\cdotpm2/C2
(a) Electric potential on the sphere surface
For a conducting sphere, the potential at its surface is: V=kQrV = \dfrac{k Q}{r}V=rkQ
Substitute: V=(9.0×109)(−40×10−9)0.03V = \dfrac{(9.0 \times 10^9)(-40 \times 10^{-9})}{0.03}V=0.03(9.0×109)(−40×10−9) V=−3600.03=−12,000 VV = \dfrac{-360}{0.03} = -12{,}000 \, \text{V}V=0.03−360=−12,000V V=−1.2×104 V\boxed{ V = -1.2 \times 10^4 \, \text{V} }V=−1.2×104V
(b) Electric potential at 6.0 cm from center
At distance r2=6.0 cm=0.06 mr_2 = 6.0 \, \text{cm} = 0.06 \, \text{m}r2=6.0cm=0.06m, the sphere behaves like a point charge: V=kQr2V = \dfrac{k Q}{r_2}V=r2kQ V=(9.0×109)(−40×10−9)0.06V = \dfrac{(9.0 \times 10^9)(-40 \times 10^{-9})}{0.06}V=0.06(9.0×109)(−40×10−9) V=−3600.06=−6,000 VV = \dfrac{-360}{0.06} = -6{,}000 \, \text{V}V=0.06−360=−6,000V V=−6.0×103 V\boxed{ V = -6.0 \times 10^3 \, \text{V} }V=−6.0×103V
Explanation (≈300 words)
The electric potential due to a charged conducting sphere depends on the distribution of charge and the distance from its center. For a conducting sphere, all the charge resides on its outer surface because like charges repel and move as far apart as possible. This makes the sphere behave as if all its charge were concentrated at the center when calculating potential outside the sphere.
The formula for electric potential at a distance rrr from a point charge is: V=kQrV = \dfrac{k Q}{r}V=rkQ
where k=9.0×109 N\cdotpm2/C2k = 9.0 \times 10^9 \, \text{N·m}^2/\text{C}^2k=9.0×109N\cdotpm2/C2, QQQ is the charge, and rrr is the distance from the center.
Inside the conducting sphere, the potential is constant and equal to the potential at the surface. Therefore, to calculate the potential at the surface, we substitute the sphere’s radius into the formula. With a radius of 0.03 m and a charge of -40 nC, the potential at the surface is: V=−1.2×104 VV = -1.2 \times 10^4 \, \text{V}V=−1.2×104V
The negative sign indicates that the potential is lower than zero since the charge is negative.
For the point outside the sphere at 6.0 cm, the same formula applies because the sphere acts like a point charge beyond its radius. Using r=0.06 mr = 0.06 \, \text{m}r=0.06m, we find the potential to be: V=−6.0×103 VV = -6.0 \times 10^3 \, \text{V}V=−6.0×103V
This shows that the potential decreases as the distance increases, following an inverse relationship with distance. The concept is crucial in electrostatics because it explains how charged objects influence their surroundings. These calculations illustrate how potential varies with distance and why conductors maintain uniform potential on their surface.