Two charges Q1=2.8 nC and Q2=-2.8 nC are separated by 11 mm. Calculate the electric potential at point B.
a. 32 kv
b. 2000 v
c. 3.5 kv
d. -250 v
Share
Two charges Q1=2.8 nC and Q2=-2.8 nC are separated by 11 mm. Calculate the electric potential at point B.
a. 32 kv
b. 2000 v
c. 3.5 kv
d. -250 v
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Answer:
The electric potential at point B due to a point charge Q can be calculated by the formula:
V = kQ/r
where k is the Coulomb constant (9 x 10^9 N m^2/C^2) Q is the charge, and r is the distance from the charge to the point B
First, let's calculate the electric potential at point B due to Q1:
V1 = kQ1/r1 = (9 x 10^9 N m^2/C^2) x (2.8 x 10^-9 C)/(0.011 m) = 254.545 V
Note that we use the magnitude of the charge since we are only interested in the electric potential, not the direction.
Next, let's calculate the electric potential at point B due to Q2:
V2 = kQ2/r2 = (9 x 10^9 N m^2/C^2) x (2.8 x 10^-9 C)/(0.011 m) = -254.545 V
Note that the negative sign indicates that the electric potential due to Q2 is in the opposite direction to that of Q1.
The total electric potential at point B is the sum of V1 and V2:
V = V1 + V2 = 254.545 V - 254.545 V = 0 V
Therefore, the answer is not given in the options provided.