Basic physics help! A charge of -2.5 × 10-9 C is at the origin and a charge of 7.9 × 10-9 C is on the x-axis a?
A charge of -2.5 × 10-9 C is at the origin and a charge of 7.9 × 10-9 C is on the x-axis at x = 3 m. At what location on the x-axis is the electric field zero?
Can't figure it out to save my life. Got to doing some quadratic formula stuff and still couldn't get it.
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Between the charges, the fields add because they are opposite in sign. The fields will therefore never cancel between the charges. The fields will only cancel outside the charges. Let x be the distance to the left of the origin:
E = kq/r^2. Factor out the 10^-9 since it appear in both terms, likewise with k
k*1e-9[2.5/x^2 - 7.9/(3+x)^2] = 0
Re-arranging:
5.4x^2-15x-22.5=0
A = 5.4, B = -15, C = -22.5
Use the quadratic formula:
B/2A +/- sqrt[B^2-4*A*C]/2A
x = 3.85783 which is the distance to the LEFT of the origin so the coordinates will actually be
(-3.85783,0)
Checking to the right of x=3 yields only negative numbers meaning the fields don't cancel for x > 3.
Naturally the field will decrease to zero far enough away from the charges. That is not the solution being sought obviously
making use of the formulation E = (a million/4piE°)(q/d^2) the place E° is absoute permitivity of loose area d= 0.25m q = 6.0 × 10–9 C (a million/4piE°)=(9*10^9) hence electric powered container= (9*10^9)*(6*10^-9)/(0.25)^2 = 864 N/C