The potential in a region between x = 0 and x = 6.00 m is V = a + bx , where a =

Question

The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 11.2 V and b = -7.10 V/m.

(a) Determine the potential at x = 0.
V

Determine the potential at x = 3.00 m.
V

Determine the potential at x = 6.00 m.
V

(b) Determine the magnitude and direction of the electric field at x = 0.

Determine the magnitude and direction of the electric field at x = 3.00 m.

Determine the magnitude and direction of the electric field at x = 6.00 m.

(a) What is the net force exerted by the two 2.20-C charges on the charge q?
N

(b) What is the electric field at the origin due to the two 2.20-C particles?
N/C

(c) What is the electrical potential at the origin due to the two 2.20-C particles?
kV

The two charges in the figure below are separated by d = 3.50 cm. (Let q1 = -13.5 nC and q2 = 25.5 nC.)

(a) Find the electric potential at point A.
kV

(b) Find the electric potential at point B, which is halfway between the charges.
kV

the two charges in the figure below are separated by a distance d = 2.50 cm, and Q = +6.70 nC.

(a) Find the electric potential at A.
kV

(b) Find the electric potential at B.
kV

(c) Find the electric potential difference between B and A.
V

A spherical conductor has a radius of 14.0 cm and a charge of 42.0 µC. Calculate the electric field and the electric potential at the following distances from the center.

(a) r = 6.0 cm

(b) r = 28.0 cm

(c) r = 14.0 cm

Explanation / Answer

Q1.The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 11.2 V and b = -7.10 V/m.

i. the potential at x = 0 is V= a + bx0 = a =11.2 V

ii. the potential at x = 3.00 m is V= a + b x 3 = 11.2 + 3 x (-7.1) = 11.2 -21.3 = -9.1 V

iii. the potential at x=6.00 m is V = a + b x 6 = 11.2 + 6 x (-7.1) = 11.2 - 42.6 = -31.4 V

Q2. the magnitude and direction of the electric field E =-dV/dx= -b =7.1 V/m

the magnitude and direction of the electric field at x = 0.0 m =7.1 V/m ..field is -x to +x

the magnitude and direction of the electric field at x = 3.0 m =7.1 V/m ..field is -x to +x

the magnitude and direction of the electric field at x = 6.0 m =7.1 V/m ..field is -x to +x

all the best in the course work

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