i think given options are
wrong... If we divide 0.01125J by 0.2m then we can get 0.056J .
You may choose laat option .
P and Q are points within a uniform electric field that are separated by a distance...
15. Point P and point charge Q are separated by a distance R. The electric field at point Phas magnitude E. How could the magnitude of the electric field at point P be doubled to 25? O double the distance to 2R O reduce the distance to R2 reduce the distance to R/4 reduce the point charge to Q2 double the point charge to 20 11. What is the electric field strength atx-0 m, y-0 m produced by two electrons,...
A point charge q = +39.0 μC
moves from A to B separated by a distance d = 0.171 m in the
presence of an external electric field E with arrow of magnitude
300 N/C directed toward the right as in the following figure.
4. +-22 points SerCP11 16 P.006 My Notes Ask Your Teacher a distance d -0.171 m in the presence of an external electric field E of magnitude 300 N/C directed toward the right as in A...
A point charge q = +39.0 µC moves from A to B separated by a distance d = 0.184 m in the presence of an external electric field E with arrow of magnitude 295 N/C directed toward the right as in the following figure. A positive point charge q is initially at point A, then moves a distance d to the right to point B. Electric field vector E points to the right. (a) Find the electric force exerted on...
A point charge q = +41.0 µC moves from A to B separated by a distance d = 0.167 m in the presence of an external electric field E with arrow of magnitude 265 N/C directed toward the right as in the following figure. A positive point charge q is initially at point A, then moves a distance d to the right to point B. Electric field vector E points to the right. (a) Find the electric force exerted on...
A point charge of −5.7 μC is located in a uniform electric field of 31 N/C. You move the charge, at constant speed, through a distance of 0.74 m exactly opposite to the direction of the field. What is the resulting change in the electric potential energy of the system?
1. A particle with a charge of +4.20 nC is in a uniform electric field E⃗ directed to the negative x direction. It is released from rest, and after it has moved 6.00 cm , its kinetic energy is found to be 1.50×10−6 J . a. What work was done by the electric force? b. What was the change in electric potential over the distance that the charge moved? c. What is the magnitude of E? d. What was the change...
A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 249 N/C. The electrical potential energy of the charge decreases by 117.7 × 10−19 J as it moves. Find the magnitude of the charge on the moving particle. The electrical potential energy depends on the distance moved in the direction of the field. Answer in units of C.
A uniform electric field of magnitude 250 V/m is directed in the positive x direction. A +11.0 μC charge moves from the origin to the point (x, y) = (20.0 cm, 50.0 cm). (a) What is the change in the potential energy of the charge field system? (b) Through what potential difference does the charge move?
A uniform electric field of magnitude E = (180) V/m is pointed in the positive x-direction. A positive charge, Q = + (11) μC (micro-coulombs) moves from an origin (0, 0) to a point (x, y) = (13, 28) cm. Through what potential difference, in volts (V) does the charge move? Round your answer to two significant figures.
A uniform electric field of magnitude 211 V/m is directed in the positive x direction. A -13.5 μC charge moves from the origin to the point (x, y) = (21.7 cm, 54.1 cm). What was the change in the potential energy of this charge? Through what potential difference did the charge move?