Question

Three charged metal disks are arranged as shown (cutaway view). The disks are held apart by insulating supports not shown in the diagram. Each disk has an area of 2.6 m² (this is the area of one flat surface of the disk). Use the value 8.85e-12 C²/(N

Answer 1

1) Disk 1 and 2 are acting like a parallel plate capacitor so the Electric filed E₁₂ between the two disks can be calculated as

$E_{12}= \frac{\sigma}{\epsilon_0}$

Here $\sigma$ is the surface chare density and is equal to charge per unit area. The above expression can be rewritten as

$|E_{12}|= \frac{Q_1}{A \epsilon_0}= \frac{5 \times 10^{-8} C}{(2.6 m^2) (8.85 \times 10^{-12} C^2/N-m^2)} =2.2 \times 10^3 N/C$

The magnitude of electric field is 2.2 x 10³ N/C.

2) The electric field is directed from disk 1 to disk 2 or from B to C. As the field lines come out of positively charged plates and enters in negatively charged plate.

3) Usually the potential diference between the two points is written as V_final - V_initial . So to describe V_C- V_Bwe are choosing the path direction that starts from B and ends at C that is option B.

4) For the considered path the sign should be negative. The directions of electric field and the $\Delta x$ (direction of the chosen path) are same and negative sign is obtained due to the following formula

$V_C-V_B= -E \Delta x$

5) The potential differece V_C-V_B is given as

$V_C-V_B= -(2.2 \times 10^3 N/C)(4 \times 10^{-3}m)=-8.8V$

6) Electric filed inside a conductor is zero. Hence V_D-V_C is also equal to zero.

7) The magnitude of electric field between disk 2 and disk 3 is

$|E_{23}|= \frac{Q_2}{A \epsilon_0}= \frac{3 \times 10^{-7} C}{(2.6 m^2) (8.85 \times 10^{-12} C^2/N-m^2)} =1.3 \times 10^4 N/C$

8) The direction of electric field is from disk 3 to disk 2 of from F to D.

9) Here, we have to calculate V_F- V_D and in this case the direction of electric field is opposite to the path so we drop the negative sign

$V_F-V_D= (1.3 \times 10^4 N/C)(1.7 \times 10^{-3}m)= 22.1 V$

10) Electric filed inside a conductor is zero. Hence, V_G-VF =0V.

11) Potential difference is a scalar quantity. The potential difference V_G- V_B is the sum of the potential differece across V_B-V_C and the potential difference across V_D-V_F as the other values of potential difference are zero.

$V_G-V_B= 22.1 V -8.8 V= 13.3 V$

To calculate change in potential energy $\Delta P.E$

$\Delta P.E = (V_G-V_B) q= (13.3 V)(-1.6 \times 10^{19}C)=-2.13 \times 10^{-20}J$

Since the external work done on the system is zero. Hence the change in kinetic energy is

+2.13 x 10^-20 Joules.