# A 25 μC point charge and -49 μC point charge are placed along the x-axis at x = 0 m and x = 3 m, . Determine point, along the x-axis, ?

A 25 μC point charge and -49 μC point charge are placed along the x-axis at x = 0 m and x = 3 m, respectively. Determine the point, along the x-axis, at which the electric field is zero.

### 1 Answer

- billrussell42Lv 73 weeks ago
let x be location of test point, distance from 0, ie |x|

test 1, assume test point is >3 m

(test 2 would be x<3m)field due to Q1 = kQ/r² = k25/x², direction to rightfield due to Q2 = kQ/r² = k49/(x–3)², direction to leftset them equal and solve for xk25/x² = k49/(x–3)²25(x² – 6x + 9) = 49x²25x² – 150x + 225 = 49x²24x² + 150x – 225 = 0x = 5/4 or –15/2since these are not >3, try next testtest 2, assume test point is <3 m, >0field due to Q1 = kQ/r² = k25/x², direction to leftfield due to Q2 = kQ/r² = k49/(3–x)², direction to rightthese will never calceltest 2, assume test point is <0field due to Q1 = kQ/r² = k25/x², direction to leftfield due to Q2 = kQ/r² = k49/(x+3)², direction to rightset them equal and solve for xk25/x² = k49/(x+3)²25(x² + 6x + 9) = 49x²25x² + 150x + 225 = 49x²24x² – 150x – 225 = 0x = 15/2, –5/4since x is |x|, x = 15/2, and position is –15/2 or –7.5 ⬅checkfield due to Q1 = kQ/r² = k25/(7.5)² = k(0.4444444)field due to Q2 = kQ/r² = k49/(10.5)² = k(0.4444444)note that I have left out the µ factor since it would cancel out anyway.The strength or magnitude of the field at a given pointis defined as the force that would be exerted on apositive test charge of 1 coulomb placed at that point;the direction of the field is given by the direction ofthat force. E = F/Q = kQ/r² Q = F/E F = QEin Newtons/coulomb OR volts/meter k = 1/4πε₀ = 8.99e9 Nm²/C²