1. An electron and a proton have charges of an equal magnitude but opposite sign
ID: 2128315 • Letter: 1
Question
1. An electron and a proton have charges of an equal magnitude but opposite sign of 1.60 10-19 C. If the electron and proton in a hydrogen atom are separated by a distance of 8.00 10-11 m, what are the magnitude and direction of the electrostatic force exerted on the electron by the proton?
magnitude ________ N
direction--> attractive/repulsive??
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2. A test charge of +8.00 10-6 C experiences a downward electrostatic force of 11 N when placed at a certain point in space. What are the magnitude and direction of the electric field at this point?
magnitude _______ N
direction up/down??
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3. Two positive charges, each of magnitude 3. 10-6 C, are located a distance of 4 cm from each other.
(a) What is the magnitude of the force exerted on each charge?
_______ N
(b) On a drawing, indicate the directions of the forces acting on each charge.
______________________________________________________________________________________
4. A charge of -3.50 10-6 C is placed at a point in space where the electric field is directed toward the right and has a magnitude of 7.50 104 N/C. What are the magnitude and direction of the electrostatic force on this charge?
magnitude _______ N
direction right/up/left/down??
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5. Two charged particles exert an electrostatic force of 5 N on each other. What will the magnitude of the electrostatic force be if the distance between the two charges is reduced to one-half of the original distance?
___________ N
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6. A charge of +5.00 10-6 C is located 11 cm from a charge of -4.00 10-6 C.
(a) What is the magnitude of the force exerted on each charge?
_________ N
(b) On a drawing, indicate the directions of the forces acting on each charge.
Explanation / Answer
1.the force is
F = k x (q1q2/r^2)
where k = (1/4pi x e_o) = 9 x 10^9 Nm^2/C^2,q1 = q2 = 1.6 x 10^-19 C and r = 8.00 x 10^-11 m
2.the test charge is q = +8.00 x 10^-6 C
the electrostatic force is F = 11 N
we know that
F = E x q
where E is electric field
or E = (F/q)
3.the charges are q1 = q2 = 3.0 x 10^-6 C
the distance between them is r = 4 cm = 4 x 10^-2 m
the force between them is
F = k x (q1q2/r^2)
the direction of the force is repulsive
4.the charge is q = -3.50 x 10^-6 C
the electric field is E = 7.50 x 10^4 N/C
the electrostatic force is
F = E x q
the direction of the force is towards left as the charge is negative
5.the force is proportional to square of inverse of distance,that is,
F1 = k x (1/r1^2)
where k is constant
similarly,
F2 = k x (1/r2^2)
or F2/F1 = (r1/r2)^2
or F2 = F1 x (r1/r2)^2
here,r2 = r1/2
or F2 = F1 x (2)^1/2 = 5 x (2)^1/2 N
6.the charges are q1 = +5.00 x 10^-6 C
and q2 = -4.00 x 10^-6 C
they are separated by distance r = 11 cm = 11 x 10^-2 m
the electrostatic force on each charge is
F = k x (q1q2/r^2)
the direction of the force is from charge q1 to charge q2(attractive force)
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