A dipole lies on the y axis and consists of an electron at y 1 = 0.50nm and a pr

Question

A dipole lies on the y axis and consists of an electron at y1 = 0.50nm and a proton at y2 = -0.50nm .

Part A

Find the electric field midway between the two charges.

Express your answer using two significant figures. Enter your answers numerically separated by a comma.

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Part B

Find the electric field at the point x = -1.5nm , y = 0.

Express your answer using two significant figures. Enter your answers numerically separated by a comma.

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Part C

Find the electric field at the point x = -26nm , y = 0.

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Explanation / Answer

A similar question is solved below, but with different values. Please workout using your figures. Hope this helps you. Please rate me.

A dipole lies on the y axis, and consists of an electron at y = 0.50 nm and a proton at y = -0.50 nm. Find the electric field at the following locations.
a) midway between the two charges
0 N/C i + _____ N/C j

(b) at the point x = 2.2 nm, y = 0
0 N/C i + _____ N/C j

(c) at the point x = -15 nm, y = 0
0 N/C i + _____ N/C j

Answer

resultant electric field E = 2kqa(a2 + x2)-3/2j^

charge q = 1.6*10-19 C

   distance a = 0.5 nm = 0.5*10-9 m

   constant k = 9*109 N.m2/C2

a)

midway between the two charges(x = 0):

electric field E = 0 N/C i^ + 2kqa(a2 + x2)-3/2j^

                           = 0 N/C i^ + 2kqa(a2 + 02)-3/2j^

                           = 0 N/C i^ + [2kq / (a)2 ]j^

                           = 0 N/C i^ + 2(9*109 N.m2/C2)(1.6*10-19 C) /(0.5*10-9 m)2

                           = 0 N/C i^ + 11.52*109 N/C j^

b)

at the point x = 2.2 nm, y = 0
            electric field E = 2kqa((0.5*10-9 m)2 + (2.2*10-9 m)2)-3/2j^

                                     = 2kqa((0.25*10-18 m + 4.84 m*10-18 )-3/2j^

                                     = 0 N/C i^ + 0.125*109 N/C j^

c)

at the point x = -15 nm, y = 0
electric field E = 2kqa((0.5*10-9 m)2 + (-15*10-9 m)2)-3/2j^                         = 2kqa((0.25*10-18 m + 225 m*10-18 )-3/2j^

                         = 4.26*105 N/C j^

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