Equipotential surfaces A positive point charge is surrounded by an equipotential
ID: 2219025 • Letter: E
Question
Equipotential surfaces
A positive point charge is surrounded by an equipotential surface A, which has a radius of rA. A positive test charge moves from surface A to another equipotential surface B, which has a radius rB. In the process, the electric force does negative work.
(a) Does the electric force acting on the test charge have the same or opposite direction as the displacement of the test charge?
The electric force has the same direction as the displacement of the test charge.
The electric force has a direction that is opposite to the displacement of the test charge.
(b) Is rB greater than or less than rA?
rB is greater than rA.
rB is less than rA.
The positive point charge is q = +8.0 10-8 C, and the test charge is q0 = +6.4 10-11 C. The work done by the electric force as the test charge moves from surface A to surface B is WAB = -8.1 10-9 J. The radius of surface A is rA = 1.8 m.
(c) What is the algebraic expression for the potential VB of surface B? Express your answer in terms of the potential VA of surface A, the work WAB, and the test charge q0. (Answer using V_A for VA, W for WAB, and q_0 for q0.)
(d) What is the algebraic expression for the potential V that the point charge q produces at an equipotential surface of radius r? Express your answer in terms of q, r, and the constant k. (Answer using q, r, and k.)
(e) Find rB, the radius of surface B.
Number
Unit
rB =
(No Response)
(No Response)
Equipotential surfaces
A positive point charge is surrounded by an equipotential surface A, which has a radius of rA. A positive test charge moves from surface A to another equipotential surface B, which has a radius rB. In the process, the electric force does negative work.
(a) Does the electric force acting on the test charge have the same or opposite direction as the displacement of the test charge?
The electric force has the same direction as the displacement of the test charge.
The electric force has a direction that is opposite to the displacement of the test charge.
(b) Is rB greater than or less than rA?
rB is greater than rA.
rB is less than rA.
The positive point charge is q = +8.0 10-8 C, and the test charge is q0 = +6.4 10-11 C. The work done by the electric force as the test charge moves from surface A to surface B is WAB = -8.1 10-9 J. The radius of surface A is rA = 1.8 m.
(c) What is the algebraic expression for the potential VB of surface B? Express your answer in terms of the potential VA of surface A, the work WAB, and the test charge q0. (Answer using V_A for VA, W for WAB, and q_0 for q0.)
(d) What is the algebraic expression for the potential V that the point charge q produces at an equipotential surface of radius r? Express your answer in terms of q, r, and the constant k. (Answer using q, r, and k.)
(e) Find rB, the radius of surface B.
Number
Unit
rB =
(No Response)
(No Response)
Please SHOW WORK I AM ATTEMPTING TO LEARN THIS THANK YOU SO MUCH!!
Explanation / Answer
1--The electric force has a direction that is opposite to the displacement of the test charge.
2--rB is less than rA.
Equipotential surface A having radius = ra has potential Va and Equipotential surface B having radius = rb has potential Vb
Wab = Work done by the Electric field = - 8.1*10^(- 9) J .
Test charge = qo = + 4.5*10^(- 11) C
Wab / qo = Va - Vb
=> - {8.1*10^(- 9)} / {4.5*10^(- 11)} = Va - Vb
=> Va - Vb = - 1.8*10² (J/C) = -180 Volt
=> Vb - Va = 180 Volt
=>k Q { (1/rb ) - (1/ra)}
Potential due to the charge producing the Electric field = V = k(Q/r)
where k = 1/4o = 9*10^9 N-C²/ m²
Q = Charge producing the electric field = + 7.2*10^(- 8) C ; ra = 1.8 m
Va = kQ/ra = (9*10^9){7.2*10^(- 8)}/1.8 = 360 volt
Vb = Va + 180 = 360 + 180 = 540 Volt
Vb = kQ/rb = (9*10^9){7.2*10^(- 8)}/rb = 540
=> rb = (9*10^9){7.2*10^(- 8)}/540 = 648 / 540 = 1.2 m
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