# How to do a and b?

Two particles having charge of +0.540 nC and +8.64 nC are separated by a distance of 1.90 m.

(a) At what point along the line connecting the two charges is the net electric field due to the two charges equal to zero?

(b) Where would the net electric field be zero if one of the charges were negative?

Relevance
• (a)

Lets consider a point P where the electric field is zero, at 'd' metres from +0.540 nC charge

Step 1: Find the direction of electric field.

Since both charges are positive the electric field will be pointed away from both charges. Let's consider direction from +0.540 nC charge to P as positive, then for the other charge it will be negative

Step 2: Calculate magnitudeElectric field due to + 0.540 nC

E₁ = kQ₁/d²

E₁ =(9 x 10⁹ Nm²/C²)(+5.40 x 10⁻¹⁰ C)/d²

E₁ = 4.86 /d² ...............(1)

Electric field due to + 8.64 nC

E₂ = -kQ₂/(1.90 - d)²

E₂ = -(9 x 10⁹ Nm²/C²)(+8.64 x 10⁻⁹ C)/(1.90 - d)²

E₂ = - 77.76 /(1.90 - d)² ...............(2)

At point P

E(net) = E₁ + E₂ = 0

4.86 /d² - 77.76 /(1.90 - d)² = 0

4.86(1.90 - d)² - 77.76d² = 0

4.86(3.61 - 3.80d + d²) - 77.76d² = 0

17.5 - 18.5d + 4.86d² - 77.76d² = 0

-72.84d² - 18.5d + 17.5 = 0

72.84d² +18.5d - 17.5 = 0

d = {-18.5) ±√[(18.5)² - 4(72.84)(-17.5)]}/(2*72.84)

d = (-18.5 ±73.76)/(2*72.84)

d = 0.379

( ignore the negative value as it will not be in between the charges)

Answer: The electric field will be 0 at 0.379 m from +0.540 nC charge

(b) If one of the charges is negative the the electric field at any point in between the charges will be directed towards the negative charge. So the net electric field will be 0 somewhere beyond the negative charge