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Course Content O waiting for response f × + v --> ) www.webassign.netweb/Student/Assignment.Respon Display in a New Window Explore A charge of Q-3.80 x10 C is located at the origin. (A) Find the magnitude of the electric force acting on a test distance of 1 m (1.00 x 10 6 m) from the origin. harge at, with charge 1.60 0aCi at- (B) Calculate the magnitude of the electric fiald at this same paint when the test charge is removed. (C) What charge in place of Q would produce this same field at a distance f- 2 times as far from the origi Conceptualize The electric field E produced by a charge Q at the location of a small "test' charge go is defined as the electric force F exerted by Q on go divided by the test charge 4 is a problem that explores the connection between the electric force and the alectric field, as well as the relationship betwean the electric field and the distance (A) Find the magnitude of the electric force acting on a test charge qo. with charge 1.6 x 101 C, at a distance of 1 m (1.0 x 10-m) from from the charge creating the field. Analyze the origin. The magnitude of the electric force between two charges is given by Therefore, the force a distance 1.0 m from Q is #x 8.99 x 10° N·m2/c" (3.80 × 10 s,C)(1.60 3.80 x 1019 C)(1.60 x 1019 c 10', c) F 8.99 x 10 N m/c =15 47 16 N. 1.0×104 m (e) Calculate the magnitude of the electric field at this same point when the test charge is removed. The field due to a charge Q at the position of go is given by magnitude of the electric Therefore, the electnc field a distance 1.0 m from O 3.80 x 100 c 34102 N/C. 1.0 x 10 m (C) What charge in place of Q would produce this same field at a distance f- 2 times as far from the origin? Replace r by fr in the equation for the electric field from the modified charge Qnes to obtain the new fiold at the changed distance. Set this equal to the previous field value to obtain which yields C. Finalize Notice how the definition of the electric field makes it possible to independent of the charge on the test particle. sign fiald values to lscations in space, 0 Type here to search

Explanation / Answer

Ans

We are given that   Qnew = f2 Q and f = 2

Therefore Qnew = 4xQ = 4x3.80x10-19 = 1.52x10-18 C

Qnew = 1.52x10-18 C

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