Question

the electric field at (30,30)cm due to a charge of -8 nano coulomb at theorigin in N/C is ...
a)-400(i+j)
b)400(i+j)
c)-200√2(i+j)
d)200√2(i+j).
plz explain with clear solution...

Answer 1

Distance of the charge from the origin = d=$\sqrt{0.30^2+0.30^2}=0.30*\sqrt2\ cm\\\\E = \frac{1}{4 \pi \epsilon_0} * \frac{q}{d^2}\\\\=9*10^9*(-8 * 10^{-9})*1/(0.30^2*2)\ N/C\\\\E=-400\ N/Coulomb$

The electric field strength is in the direction of  Ф   where
     Tan Ф = 30 cm/ 30 cm = 1    => 45 deg. with X- axis.
      Cos Ф = 1/√2  and Sin Ф = 1/√2
Since E is negative, the Ф is in the 3th quadrant.  ie., Ф = 180+45 deg. = 225 deg.

The vector form of electric field = $\vec{E}=E [ Cos\phi \hat{i}+ sin\phi \hat{j} ]\\\\\vec{E}= - (400/\sqrt2) [ \hat{i}+\hat{j}] Newtons/Coulomb\\\\\vec{E}= -200\sqrt2 (\hat{i}+\hat{j}] Newtons/Coulomb$