Concepts Of Physics MCQ Edition [Volume 2]PhysicsElectric Field and Potential
Certain equipotential surfaces are depicted in the figure. Determine the magnitude and direction of the electric field in both situations.
Options
- A(a) 200 V m ⁻¹ making an angle of 60^ with the x -axis; (b) radially inward, decreasing as E = 6 V m r^2
- B(a) 100 V m ⁻¹ making an angle of 60^ with the x -axis; (b) radially inward, decreasing as E = 600 V m r^2
- C(a) 200 V m ⁻¹ making an angle of 120^ with the x -axis; (b) radially outward, decreasing as E = 6 V m r^2
- D(a) 100 V m ⁻¹ making an angle of 120^ with the x -axis; (b) radially outward, decreasing as E = 600 V m r^2
Correct answer
C. (a) 200 V m ⁻¹ making an angle of 120^ with the x -axis; (b) radially outward, decreasing as E = 6 V m r^2
Step-by-step solution
In situation (a), the equipotential lines are parallel, indicating a uniform electric field. The perpendicular distance r between the 10 V and 20 V lines is r = (20 cm - 10 cm ) 30^ = 10 0.5 = 5 cm = 0.05 m . Magnitude of electric field E = V r = 20 - 10 0.05 = 200 V m ⁻¹ . The electric field is perpendicular to the equipotential lines and points towards decreasing potential (negative x-direction). Thus, the angle with the positive x-axis is 30^ + 90^ = 120^ . In situation (b), the equipotential surfaces are concen