Physics Self-Test - Page 15

108- The total capacitance of capacitors connected in parallel is given by _________
Explanation: For a parallel array configuration, the total equivalent capacitance is simply the direct algebraic sum of individual values: $C_{eq} = C_1 + C_2 + C_3 + \dots$ because all plates share the identical potential drop, expanding the collective effective surface area.
109- What is the total charge on the parallel plate capacitor shown?
Q -Q
Explanation: While the magnitude of charge stored on an individual isolated plate is $Q$, a capacitor as an integrated component holds equal and opposite internal charges ($+Q$ and $-Q$). Summing up the entire component yield evaluates to zero ($Q + (-Q) = 0$).
What happens to the capacitance when a dielectric material is inserted between the plates of a parallel plate capacitor?
Explanation: Adding a dielectric material reduces the effective internal electric field via dielectric polarization by a relative factor of $\kappa$. Since $C = \frac{Q}{V}$ and voltage drops ($V = \frac{V_0}{\kappa}$), the capacitance scales up proportionally: $C = \kappa C_0$.
A particle having mass 1 g and electric charge $10^{-8}\text{ C}$ travels from a point A having electric potential 600 V to a point B having zero potential. What is the change in its kinetic energy?
Explanation: By conservation of mechanical energy, $\Delta K = -\Delta U = -q(V_B - V_A)$. Substituting values gives: $\Delta K = -10^{-8}\text{ C} \times (0 - 600\text{ V}) = 6 \times 10^{-6}\text{ J}$.
A moving electron approaches another electron. What would be the change in the potential energy of this system?
Explanation: Electrons are like-charged objects that naturally repel each other via Coulomb's law. Forcing them closer against this repulsive force performs work on the system, boosting its electrostatic potential energy ($U = \frac{k q_1 q_2}{r}$ as separation distance $r$ falls).