Charge qsub1 = 6.95e-6 C is at the origin, and charge qsub2 = -5.05e-6 C is on t
ID: 2149573 • Letter: C
Question
Charge qsub1 = 6.95e-6 C is at the origin, and charge qsub2 = -5.05e-6 C is on the x axis, 0.300 m from the origin (see figure).
magnitude (N/C)
direction(degrees)
Explanation / Answer
Ans a.given Electric field at P q1 = 6.95x10-6 q2 = -5.05x10-6 ==>E1 due to positive charge E2 == due to negative chagrge E1 = kq1/r^2 North E2 = kq2/r^2 Nothe west Angle = tan-1(4/3) Hence resolving horizontal and vertical components we have E2 cos x = East E2sin x = West Hence we have E = V(E1 + E2) =>E2cosx i +E1-E2sinxj =>181.3i + 160.1j x10^4 = 2.418 x 10^6 N/C Angle = 41.44 degree b.Force = E*Q = 4.836N c.It Remains Same at P and negarive charge location E1 = 3.909x10^4 N/c East d.Force on q2 => =>kq3/0.5^2 +kq1/0.3^2 ==>8.749 x 10^4i -5.65*10^4j Hence Field =.1035^N/C Direction = tan -1(0.672)