An insulating sheet in the xz-plane is uniformly charged with charge surface density $\sigma$ = 6.2 * 10^-6 C/m².What is the change in electric potential when a charge of Q=3.3 $\mu$C is moved from a position that is 6.2 m away from the sheet to a position that is 1.3 m away from the sheet?

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Quizplus · Accepted Answer

The electric potential due to an infinite sheet of charge is given by $V = \frac{\sigma}{2\epsilon_0}z$, where $\sigma$ is the surface charge density, $\epsilon_0$ is the permittivity of free space, and $z$ is the distance from the sheet. The change in potential, $\Delta V$, when moving a charge $Q$ from a distance $z_1$ to $z_2$ is $\Delta V = V_2 - V_1 = \frac{\sigma}{2\epsilon_0}(z_2 - z_1)$. Since the potential does not depend on the value of $Q$, and given that $\sigma = 6.2 \times 10^{-6} \, \text{C/m}^2$, $z_1 = 6.2 \, \text{m}$, and $z_2 = 1.3 \, \text{m}$, the change in potential is determined by the difference in distances and the surface charge density, leading to a positive change in potential due to moving closer to the sheet.

An Insulating Sheet in the Xz-Plane Is Uniformly Charged with Charge