Question

Two point charges 3 microcoulomb and 2.5 micro coulomb are placed at point a 1, 1, 2 metre and B 0, 3, - 1 metre respectivelty find out electric field intensity at a point c 3, 3, 3meter

Answer 1

Answer:

Two point charges 3 microcoulomb and 2.5 micro coulomb are placed at point a 1, 1, 2 metre and B 0, 3, - 1 metre respectivelty

The electric field intensity at the point C(3,3,3)C3,3,3

Due to Charge at A is,

E→1=1/4πε03×10^(−6)/r^3  r→

E→1=1/4πε03×10^(−6)/√((3−1)^2+(3−1)^2+(3−2)^2) ^3((3−1) iˆ + (3−1) jˆ + (3−2) kˆ)

E→1=9×10^9×3×10^(−6)/√(4+4+1)^3(2 iˆ + 2 jˆ + kˆ)

E→1=27×10^3/27(2 iˆ + 2 jˆ + kˆ)

E→1=(2 iˆ + 2 jˆ + kˆ)×10^3 N/C

Due to charge at B is,

E→2=1/4πε02.5×10^(−6)/r^3r→

E→2=1/4πε02.5×10^(−6)/√((3−0)^2+(3−3)^2+(3+1)^2)^3((3−0) iˆ + (3−3) jˆ(3+1)kˆ)

E→2=9×10^9×2.5×10^(−6)/√(9+16)^3(3 iˆ +4 kˆ)

E→2=22.5×10^3/125(3 iˆ +4 kˆ)

E→2=(3 iˆ +4 kˆ)×0.18×10^3 N/C

Therefore, according to the superposition principle, the net electric field intensity at C is,

E→=E→1+E→2

=(2 iˆ + 2 jˆ + kˆ)×10^3+(3 iˆ + kˆ)×0.18×10^3

E→=[(2+0.54) iˆ + (2+0) jˆ +(1+0.18) kˆ] ×10^3

E→=[2.54 iˆ +2 jˆ +1.18 kˆ] kN/C

∣E→∣=√(2.54)2+(2)2+(1.18)^2 kN/C

∣E→∣= √6.45+4+1.39 kN/C

∣E→∣=√11.844−−−−−− kN/C

∣E→∣=3.44 kN/C