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Question 11

Topic: Coulomb Force Between Parallel Plates

• Two parallel metal plates are separated by a distance of 0.5 cm and have a potential difference of 10,000 V. The lower plate is negatively charged, and the upper plate is positively charged. You are asked to find the magnitude and direction of the Coulomb force on an electron placed between the plates.

To solve this:

1. Use the formula for electric field $E$ between parallel plates: $E = \frac{V}{d}$ Where:

   • $V = 10,000 \, \text{V}$ is the potential difference.
   • $d = 0.5 \, \text{cm} = 0.005 \, \text{m}$ is the distance between plates.

   Substituting the values: $E = \frac{10,000}{0.005} = 2 \times 10^6 \, \text{V/m}$

2. The Coulomb force $F$ on an electron is given by: $F = qE$ Where:

   • $q = -1.6 \times 10^{-19} \, \text{C}$ is the charge of an electron.

   Substituting the values: $F = (-1.6 \times 10^{-19})(2 \times 10^6) = -3.2 \times 10^{-13} \, \text{N}$ The negative sign indicates that the force is directed towards the positively charged upper plate (opposite the electron’s charge).

Question 12

Topic: Electron Acceleration in an Electric Field

• An electron (mass $9.11 \times 10^{-31} \, \text{kg}$) is accelerated from rest between a cathode and an anode 2 cm apart with a potential difference of 200 V. Find the electron's velocity when it reaches the anode.

To solve this:

1. The energy gained by the electron is the electric potential energy, which is converted into kinetic energy: $eV = \frac{1}{2}mv^2$ Where:

   • $e = 1.6 \times 10^{-19} \, \text{C}$
   • $V = 200 \, \text{V}$
   • $m = 9.11 \times 10^{-31} \, \text{kg}$

2. Rearranging for $v$: $v = \sqrt{\frac{2eV}{m}}$

   Substituting the values: $v = \sqrt{\frac{2(1.6 \times 10^{-19})(200)}{9.11 \times 10^{-31}}} = \sqrt{\frac{6.4 \times 10^{-17}}{9.11 \times 10^{-31}}} = \sqrt{7.02 \times 10^{13}}$

   Thus: $v \approx 8.38 \times 10^6 \, \text{m/s}$

Question 13

Topic: Energy Stored in Capacitors

• Three identical uncharged capacitors are connected to a 15 V battery. If only one is initially connected to the battery, the energy stored in it is $\frac{3}{2}E$. Find the energy stored when all three capacitors are connected in series.

The energy stored in a capacitor is given by: $U = \frac{1}{2}CV^2$ When three capacitors are connected in series, the equivalent capacitance $C_{\text{eq}}$ is: $C_{\text{eq}} = \frac{C}{3}$ The energy stored in the series combination becomes: $U_{\text{series}} = \frac{1}{2} C_{\text{eq}} V^2 = \frac{1}{2} \left(\frac{C}{3}\right) V^2 = \frac{1}{6}CV^2$ Thus, the total energy stored when connected in series is $\frac{1}{6}$ the energy stored in one capacitor.

Question 14

Topic: Current in a Single Loop Circuit

• The circuit consists of two resistors (2Ω and 3Ω) connected in series to a 12V battery. You are asked to find the total current flowing through the circuit.

The total resistance in the circuit is: [ R_{\text{total