18. [1pt] Diagrams A and B show the electric potential along the x-axis in two d
ID: 1429846 • Letter: 1
Question
18. [1pt]
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.
a.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.
b.The electric potential at x = 4 m in Diag. A is larger than the electric potential at x = 5 m in Diag. B.
c.A negative charge placed at x = 4 m in Diag. A and released will accelerate left.
d.The work done by you to bring a positive charge from infinity to the point x = 1 m in Diag. A is greater than zero.
e.The electric field at x = 2 m in Diag. A points to the left.
f.The magnitude of the electric field at x = 2 m in Diag. A is larger than the magnitude of the electric field at x = 2 m in Diag. B.
Answer:
19. [1pt] Calculate a numeric value for the electric field at the point x = 2 m in diagram B.
Answer:
20. [1pt]
Calculate the force on a positive charge of 1 µC located at the point x = 4 m in diagram A. (If the force is parallel to the x axis, then the force is negative.)
Answer:
21. [1pt] Calculate the initial acceleration of a proton placed at the point
x = 4 m in diagram B. (If the acceleration is parallel to the x axis, then the acceleration is negative.)
Answer:
Explanation / Answer
(18)
(i) the HCl molecule has no charge. so, the work done is zero.-----F
(ii)
electric field at any given location is equal; to the slope of the line at that location. thus, the slope of the line in diag B is more than in diag A at x=4 m. thus, the answer is -----F
(iii)
the positive charge always accelerates towards low potential region. from x=2 m to x=3 m, the potential is increasing in diag A. so, the answer is -------F
(iv)
electric field points from high potential to low potential region. answer is---------T
(v) true.
(vi) false
----------------------------------------------
(19)
the electric field at x=2 m in figure B is slope of the line.
E= (4.0 V-2.5 V) / (3 m-1 m) = 0.75 V/m
--------------
(20)
the field at x =4 m in figure A is,
E= (4 V-3 V) /(5 m-3 m) =0.5 V/m
the required force is,
F=Eq
=(0.5 V/m)(1*10^-6 C)=5*10^-7 N
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