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Two particles are fixed to an x axis: particle 1 of charge -6.60 × 10^-7 C is at the origin and particle 2 of charge +6.60 × 10^-7 C is at?
Two particles are fixed to an x axis: particle 1 of charge -6.60 × 10^-7 C is at the origin and particle 2 of charge +6.60 × 10^-7 C is at x2 = 17.5 cm. Midway between the particles, what is the magnitude of the net electric field?
1 Answer
- nyphdinmdLv 72 months ago
q1 = -6.6x10^-7 C, q2 = 6.6x10^-7 C --> q1 = -q2 x1 = 0.175 m
field due to q1 at position x: E1(x) = kq1/x^2
field due to q2 at position x: E2(x) = kq2/(x - x1)^2 = -kq1/(x - x1)^2
Now the field from q2 points toward q1 --> points in the -x direction and the field from q1 also is pointing in the -x direction when looking toward q2. Set x = x1/2 since the distance between the charges is x1, then
E(x1/2) = kq1/(x1/2)^2 - kq2/(x1/2)^2 = -2k*|q1|/(x1/2)^2 = -8*k*|q1|/x1^2
plug in noting |q1| is the absolute value.