The Setup
A Sphere and a Charge
Imagine a perfectly neutral conducting sphere of radius R=10 cm. Sitting quietly at a distance l=20 cm from its center is a positive point charge q=10−8 C. This seemingly simple setup is a classic playground for exploring the profound properties of conductors in electrostatics.
Phase 1
Potential Before Grounding
The beauty of conductors lies in their equipotential nature. Because charges are free to move, they instantly rearrange themselves until the electric field inside the conductor is zero. This means the entire volume of the sphere, including its surface, is at the exact same potential.
To find this potential, we can pick any point on or inside the sphere. The smartest choice? The center.
At the center, the potential is the sum of the potential due to the external point charge and the potential due to the induced charges on the surface. However, because the sphere is initially neutral, the total induced charge is zero. Since all these induced charges are at the same distance R from the center, their net contribution to the potential at the center is exactly zero!
Therefore, the potential of the sphere is simply the potential created by the point charge at the center:
Vsphere=lkq=0.29×109×10−8=450 V
This confirms that statement (A) is absolutely correct.
Phase 2
The Grounding Connection
Now, we introduce a grounding wire. What does it mean to ground something? It means we connect it to the Earth, an infinitely large reservoir of charge, forcing the conductor's potential to become exactly zero.
To achieve this zero potential, charges must flow between the Earth and the sphere. Let's call the new charge on the sphere qs. We again use our center-point trick. The total potential at the center must now be zero:
Solving for qs, we get:
qs=−q(lR)=−10−8×(0.20.1)=−5×10−9 C
Phase 3
The Flow of Charge
The sphere went from having zero net charge to having −5×10−9 C. Physically, this means electrons flowed from the ground up into the sphere.
By standard convention, the direction of current (or charge flow) is taken as the flow of positive charge. If −5×10−9 C flowed to the sphere, it is mathematically identical to saying that +5×10−9 C flowed from the sphere to the ground.
This perfectly matches statement (B), making it correct.
Phase 4
Breaking the Connection
Next, we snip the grounding wire. The sphere is now isolated. The charge qs=−5×10−9 C is trapped on the sphere and has nowhere to go.
This confirms statement (C) is correct.
Final Calculation
Moving the Charge
Finally, we move the external point charge 10 cm further away. The new distance from the center is l′=20+10=30 cm=0.3 m.
Even though the charge on the sphere is trapped, the potential of the sphere will change because the external charge has moved. We calculate the new potential at the center once more:
Substituting our known values:
Vfinal=0.19×109×(−5×10−9)+0.39×109×10−8
Vfinal=−450 V+300 V=−150 V
The final potential is −150 V, not 300 V. Therefore, statement (D) is incorrect.
