Diagrams A and B show the electric potential along the x-axis in two different s

Question

Diagrams A and B show the electric potential along the x-axis in two different situations. Answer each of the questions below with True (T), False (F), or Can not tell (C). If, for example, the answer to the first question is True, the answer to the second is Can Not Tell, and the answer to the others is False, enter TCFFFF.

i. The electric potential at x = 4 m in Diag. A is larger than the electric potential at x = 5 m in Diag. B.

ii. The work done by you to bring a negative charge from infinity to the point x = 4 m in Diag. A is greater than zero.

iii. The magnitude of the electric field at x = 4 m in Diag. A is larger than the magnitude of the electric field at x = 4 m in Diag. B.

iv. A negative charge placed at x = 4 m in Diag. A and released will accelerate left.

v. The electric field at x = 2 m in Diag. A points to the left.

vi.An electric dipole with its positive charge at y = -0.01 mm and its negative charge at y = 0.01 mm placed at xx = 2 m in Diag. A and released will initially rotate counterclockwise.

Explanation / Answer

TTFTT

a) True

from the diagram A, at x = 4, V = -1V

from the diagram B, at x = 5, V = -50V

so V_a>V_b, so the given statement is true.

b) True

We know that workdone W = qV

here the charge is negative and the potential is also negative so the workdone is positive i.e. >0. So the given statement is true.

c) False

In diagram A, we have change in voltage as 2V for 2m distance where as in diagram B we have 3V for 2m distance.

We also know that E = dV/dx

i.e. the field in diagram B is larger.

d) True

at x=4m, we have a negative slope for V vs x so the electric field is positive. So when we keep the negative charge in electric field, it will accelerate opposite to the direction of motion i.e. towards left.

e) True

Here V vs x curve has positive slope so its direction is along negative axis.

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