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Concept explainers
(a)
The magnitude and direction of electric field at
(a)
![Check Mark](/static/check-mark.png)
Explanation of Solution
Given:
The magnitude of the first point charge is
The position of the first point charge is
The magnitude of the second point charge is
The position of the second point charge is
Formula used:
Write the expression for the resultant electric field at point
Here,
Write the expression for the electric field due to charge
Here,
Write the expression for the electric field due to charge
Here,
Calculation:
Substitute
Substitute
Substitute
The Magnitude of electric field is:
Direction of electric field is:
Conclusion:
The magnitude and direction of electric field is
(b)
The magnitude and direction of electric force.
(b)
![Check Mark](/static/check-mark.png)
Explanation of Solution
Given:
The magnitude of the first point charge is
The position of the first point charge is
The magnitude of the second point charge is
The position of the second point charge is
Formula used:
Write the expression for the force.
Here,
Calculation:
Substitute
The magnitude of force is:
The direction of electric force is:
Conclusion:
The magnitude and direction of force is
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Chapter 21 Solutions
Physics for Scientists and Engineers
- A point charge of 4.00 nC is located at (0, 1.00) m. What is the x component of the electric field due to the point charge at (4.00, 2.00) m? (a) 1.15 N/C (b) 0.864 N/C (c) 1.44 N/C (d) 1.15 N/C (e) 0.864 N/Carrow_forwardTwo particles with charges q1 and q2 are separated by a distance d, and each exerts an electric force on the other with magnitude FE. a. In terms of these quantities, what separation distance would cause the magnitude of the electric force to be halved? b. In terms of these quantities, what separation distance would cause the magnitude of the electric force to be doubled?arrow_forwardA 1.75-nC charged particle located at the origin is separated by a distance of 0.0825 m from a 2.88-nC charged particle located farther along the positive x axis. If the 1.75-nC particle is kept fixed at the origin, where along the positive x axis should the 2.88-nC particle be located so that the magnitude of the electrostatic force it experiences is twice as great as it was in Problem 27?arrow_forward
- What is the magnitude of the electric field due to a charged particle at its exact location (r = 0)?arrow_forwardIn nuclear fission, a nucleus of uranium-238, which contains 92 protons, can divide into two smaller spheres, each having 46 protons and a radius of 5.90 1015 m. What is the magnitude of the repulsive electric force pushing the two spheres apart?arrow_forwardIs it possible for a conducting sphere of radius 0.10 m to hold a charge of 4.0 C in air? The minimum field required to break down air and turn it into a conductor is 3.0 106 N/C.arrow_forward
- aA plastic rod of length = 24.0 cm is uniformly charged with a total charge of +12.0 C. The rod is formed into a semicircle with its center at the origin of the xy plane (Fig. P24.34). What are the magnitude and direction of the electric field at the origin? Figure P24.34arrow_forwardA rod 14.0 cm long is uniformly charged and has a total charge of 22.0 C. Determine (a) the magnitude and (b) the direction of the electric field along the axis of the rod at a point 36.0 cm from its center.arrow_forwardPanicle A of charge 3.00 104 C is at the origin, particle B of charge 6.00 104 C is at (4.00 m, 0), and panicle C of charge 1.00 104 C is at (0, 3.00 m). (a) What is the x-component of the electric force exerted by A on C? (b) What is the y-component of the force exerted by A on C? (c) Find the magnitude of the force exerted by B on C. (d) Calculate the x-component of the force exerted by B on C. (e) Calculate the y-component of the force exerted by B on C. (f) Sum the two x-components to obtain the resultant x-component of the electric force acting on C. (g) Repeat part (f) for the y-component. (h) Find the magnitude and direction of the resultant electric force acting on C.arrow_forward
- Three charged panicles are located at the corners of an equilateral triangle as shown in Figure P23.15. Calculate the total electric force on the 7.00-C charge.arrow_forwardOn a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 100 N/C. Compare the gravitational and electric forces on a small dust particle of mass 2.01015 g that carries a single electron charge. What is the acceleration (both magnitude and direction) of the dust particle?arrow_forwardPanicle A of charge 3.00 104 C is at the origin, particle B of charge 6.00 104 C is at (4.00 m, 0), and panicle C of charge 1.00 104 C is at (0, 3.00 m). (a) What is the x-component of the electric force exerted by A on C? (b) What is the y-component of the force exerted by A on C? (c) Find the magnitude of the force exerted by B on C. (d) Calculate the x-component of the force exerted by B on C. (e) Calculate the y-component of the force exerted by B on C. (f) Sum the two x-components to obtain the resultant x-component of the electric force acting on C. (g) Repeat part (f) for the y-component. (h) Find the magnitude and direction of the resultant electric force acting on C.arrow_forward
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