Problem 1.
Two small spheres spaced 35.0cm apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is ?
Solution
Problem 2.
A point charge is at the point , , and a second point charge is at the point , . Find the magnitude and direction of the net electric field at the origin.
Solution
Problem 3.
What must the charge (sign and magnitude) of a particle of mass 5 g be for it to remain stationary when placed in a downwarddirected electric field of magnitude 800 N/C?
Solution
Problem 4.
What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?
Solution
Problem 5.
A particle has a charge of 8.00 nC. Find the magnitude and direction of the electric field due to this particle at a point 0.5 m directly above it.
Solution
Problem 6.
Two particles having charges of 0.70 nC and 12 nC are separated by a distance of 2 m. At what point along the line connecting the two charges is the net electric field due to the two charges equal to zero?
Solution
Problem 7.
A closed surface encloses a net charge of 10 nC . What is the net electric flux through the surface?
Solution
Problem 8.
Each square centimeter of the surface of an infinite plane sheet of paper has excess electrons. Find the magnitude and direction of the electric field at a point 6.00 cm from the surface of the sheet, if the sheet is large enough to be treated as an infinite plane.
Solution
Problem 9.
A thin disk with a circular hole at its center, called an annulus has inner radius a nd outer radius . The disk has a uniform positive surface charge density . Find the total electric charge on the annulus.
Solution
Problem 10.
A thin disk with a circular hole at its center has inner radius a nd outer radius . The disk has a uniform positive surgace charge density on its surface. The disk lies in the yz plane, with its center at the origin. For an arbitrary point on the x axis (the axis of the disk) find the magnitude of the electric field.
Solution
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