Question

2. The electric field at (30, 30) cm due to a charge of -8 nC at the origin in NC is
1) -400 (i+j)
2) 400 ( + i)
3) -20012 +j 4) 200_2 6+
3. Two charges 4 x 10-° C and
اليا​

Answer 1

The electric field at (30, 30) cm due to a charge of -8 nC at the origin in N/C is

1. -400 (i + j)
2. 400 (i + j)
3. -200√2(i + j)
4. 200√2 (i + j)

solution : using formula of electric field in vector form, $\vec E=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$

here $\hat{r}=\frac{(30-0)\hat i+(30-0)\hat j}{\sqrt{30^2+30^2}}$

= $\frac{1}{\sqrt{2}}(\hat i+\hat j)$

here r = √{30² + 30²} = 30√2 cm

now electric field due to -8nC at (30,30) is given by, E = (9 × 10^9 × -8 × 10^-9)/(30√2 cm)² [1/√2(i + j)]

= -72/(1800 × 10¯⁴ m²) × 1/√2(i + j)

= -200√2 (i + j) N/C

Therefore the electric field due to -8nC at (30,30) is -200√2 (i + j).

option (3) is correct choice.