Please help with the following 4 questions! Please show all work so I can follow
ID: 1429090 • Letter: P
Question
Please help with the following 4 questions! Please show all work so I can follow
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 Cannot 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. 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. 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. A positive charge placed at x = 2 m in Diag. A and released will accelerate right. The electric field at x = 4 m in Diag. A points to the left. 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. The electric potential at x = 3 m in Diag. A is larger than the electric potential at x = 2 m in Diag. B. Calculate a numeric value for the electric field at the point x = 4 m in diagram A. Calculate the force on a positive charge of 1 MewC located at the point x = 4 m in diagram B. (If the force is parallel to the -x axis, then the force is negative.) Calculate the initial acceleration of a proton placed at the point x = 2 m in diagram A. (If the acceleration is parallel to the -x axis, then the acceleration is negative.)Explanation / Answer
Question 1 :
i) True .
Potential at infinity = 0
Potential at X= 4 , = -1 volts
Potential difference = -1 volts
work done = (-q) (-1) >0
ii)
False
Slope at X=2 is smaller as compared to slope at x = 2 in B
iii)
False
at X = 2 , Slope is positive and electric potential is increasing , so electric field is from right to left
a positive charge moves in the direction of electric field , hence charge moves towards left
iv)
False
at X= 4 Slope is negative and electric potential is decreasing , hence electric field is from left to right.
v)
False
the dipole rotates clockwise , since electric field at X = 2 is from right to left . so positive charge moves towards left and negative charge towards right
Vi)
true
0 > - 48