Use the following figure to answer the next problem: Two equal positive charges are placed x m apart.The equipotential lines are at 100 V interval.
-The electrical potential 2.5 cm from a point charge of Q₁ = +4.5  10^-9 C and 2.0 cm from a second charge Q₂ is 3.2 kV.Find Q₂.

Question

Use the following figure to answer the next problem:

Two equal positive charges are placed x m apart.The equipotential lines are at 100 V interval.
-The electrical potential 2.5 cm from a point charge of Q₁ = +4.5  10^-9 C and 2.0 cm from a second charge Q₂ is 3.2 kV.Find Q₂.

Quizplus · Accepted Answer

Using the formula for electrical potential: V = kQ/r We can find the value of Q₂: V = 3200 V = kQ₂/0.02 m Q₂ = V*0.02 m/k = 8.99*10^-9 C Now, we can find the electric field due to the first charge: E = kQ₁/(0.025 m)^2 = 7.2*10^6 N/C Next, we can find the potential due to the electric field at a distance of 0.02 m from the first charge: V1 = E*d = (7.2*10^6 N/C)*(0.02 m) = 144 kV So, the total potential at a distance of 2.0 cm from the second charge is: Vtotal = V1 + 3200 V = 323.2 kV We can find the charge of the second charge using the formula: Vtotal = kQ₂/0.025 m Q₂ = Vtotal*0.025 m/k = 8.96*10^-9 C Rounding to the nearest tenth, we get: Q₂ = 3.5 F1F1F1S1F1F1F10 10^-9 C

Use the Following Figure to Answer the Next Problem