Question

Calculate the electric potential at the center of the square described as follows: Four charges are placed at the corners of a 20 cm square. The particles are as follows: 10 microC at x =0, y = 0, -5 microC at x = 20, y = 0, -15 microC at x = 20, y = 20, and 20 microC at x=0 and y = 20 .

Answer 1

, the answer is 2020 kV

Step-by-step explanation

Electric potential of a point charge is

`\begin{gathered} V=(KQ)/(d) \\ where\text{ kis a constant k=9*10}^9\text{ N*}(m^2)/(C^2) \\ Q\text{ is charge , d is the distance} \end{gathered}`

so

Step 1

Diagram:

Step 2

distances:fromthe center to each charge:

:

`\begin{gathered} d=√(0.10^2+0.10^2)=√(0.02)\text{ m=0.1414m} \\ d=0.1414\text{ m} \end{gathered}`

Step 3

electric potential due to each charge.

a) Q1

`\begin{gathered} V=(KQ)/(d) \\ V=\text{9*10}^9\text{ N*}(m^2)/(C^2)*\frac{10*10^(-6)C}{0.1414\text{ m}} \\ V=636.49\text{ kV} \\ \end{gathered}`

b)Q2

`\begin{gathered} V=(KQ)/(d) \\ V=\text{9*10}^9\text{ N*}(m^2)/(C^2)*\frac{-5*10^(-6)C}{0.1414\text{ m}} \\ V=-318.24\text{ KV} \\ \end{gathered}`

c)Q3

`\begin{gathered} V=(KQ)/(d) \\ V=\text{9*10}^9\text{ N*}(m^2)/(C^2)*\frac{-15*10^(-6)C}{0.1414\text{ m}} \\ V=-954.73\text{kV} \\ \end{gathered}`

d)Q4

`\begin{gathered} V=(KQ)/(d) \\ V=\text{9*10}^9\text{ N*}(m^2)/(C^2)*\frac{20*10^(-6)C}{0.1414\text{ m}} \\ V=-954.73\text{kV} \\ V=1272.98\text{ kV} \end{gathered}`

so, the resultant electric potential are:

:

So

`\begin{gathered} v=√(1309.2^2+1591.22^2) \\ v=2060\text{ Kv} \end{gathered}`

therefore, the answer is 2020 kV

I hope this helps you