Problem 3: Three point charges each of 4 microC are placed at three corners of a square of side 20cm. Find the magnitude of force on each.

Solution:

Step: 1 Overview

See the overviews of Problem-1 and Problem-2.

Step: 2 Calculation

Since r₁, r₂ and r₃ form right-angeled triangle, so applying Pathagorous theorem to find the value of r₃:

r₃²=r₁²+r₂²
r₃²=(20cm)²+(20cm)²
r₃²=400cm²+400cm²
r₃²=800cm²
r₃=28.28cm

Given: q₁=q₂=q₃= 4microC = 4 x 10^-6C.
Force on Q₁:
Force on Q₁ due to Q₂ (F₁₂)

F₁₂=Kq₁q₂/r₁²
F₁₂=9x10⁹(4x10^-6)(4x10^-6)/(0.2)²
F₁₂=3.6N.

Force on Q₁ due to Q₃ (F₁₃)

F₁₃=Kq₁q₃/r₃²
F₁₃=9x10⁹(4x10^-6)(4x10^-6)/(0.2828)²
F₁₃=1.8N.

Resultant Force:

According to vector addition, if two vectors having magnitudes a and b are making angle X, then magnitude of resultant vector r can be found by the formula:
r²=a²+b²+abcosX.
So similarly,
F²=F₁₂²+F₁₃²+2F₁₂F₁₃cos45^o.
F²=(3.6)²+(1.8)²+2(3.6)(1.8)(0.707)
F²=25.364
F=5.04N. (Ans)

Force on Q₂:
Force on Q₂ due to Q₁ (F₂₁)

F₂₁=F₁₂=3.6N

Force on Q₂ due to Q₃ (F₂₃)

F₂₃=Kq₂q₃/r₂²
F₂₃=9x10⁹(4x10^-6)(4x10^-6)/(0.2)²
F₂₃=3.6N.

Resultant Force:

F²=F₂₁²+F₂₃²+2F₂₁F₂₃cos90^o.
F²=(3.6)²+(3.6)²+2(3.6)(3.6)(0)
F²=25.92
F=5.09N. (Ans)

Force on Q₃:
Force on Q₃ due to Q₁ (F₃₁)

F₃₁=F₁₃=1.8N

Force on Q₃ due to Q₂ (F₃₂)

F₃₂=F₂₃=3.6N

Resultant Force:

F²=F₃₁²+F₃₂²+2F₃₁F₃₂cos45^o.
F²=(1.8)²+(3.6)²+2(1.8)(3.6)(0.707)
F²=25.364
F=5.04N. (Ans)