Question

Solve 1.1. A 1C charge is originally a distance of 1m from a 0.2C charge, but is moved to a distance of 0.1 m. What is the change in electric potential energy?1. 0 J2. -9.0x10^9 J3. 1.6x10^10 J4. 9.0x10^9 J2. A 1C charge is a distance of 1m from a 0.2C charge. What is the electric potential at the position of the 0.2C charge?1. 9.0x10^9 V2. 0 V3. 4.5x10^10 V4. 1.8x10^9 V

Answer 1

Answer:

3. 1.6 x 10^10 J

Explanation:

The electric potential energy can be calculated as

$\begin{gathered} U=k\frac{q_1q_2}{r} \\ \\ Where\text{ k = 9 }\times10^9\text{ N m}^2\text{ /C}^2 \\ q_1\text{ : Charge 1} \\ q_2\text{ : Charge 2} \\ r:\text{ distance} \end{gathered}$

So, we can calculate the change in electric potential energy as

$\begin{gathered} \Delta U=U_2-U_1 \\ \\ \Delta U=(9\times10^9\text{ N m}^2\text{ /C}^2)\frac{(1\text{ C\rparen\lparen0.2 C\rparen}}{(0.1\text{ m\rparen}}-(9\times10^9\text{ N m}^2\text{ /C}^2)\frac{(1\text{ C\rparen\lparen0.2 C\rparen}}{1\text{ m}} \\ \\ \Delta U=1.8\times10^{10}J-0.18\times10^{10}J \\ \Delta U=1.62\times10^{10}J \end{gathered}$

Therefore, the answer is

3. 1.6 x 10^10 J