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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Physics
Chapter
Electrostatics

BITSAT 2024 · Physics · Electrostatics

Electric field at point (30,30,0) due to a charge of 0.008 μ C...

Electric field at point (30,30,0)(30,30,0) due to a charge of 0.008μC0.008 \mu \mathrm{C} placed at origin will be, (coordinates are in cm )

Options

  1. A8000NC−18000 \mathrm{NC}^{-1}
  2. B4000(i^+j^)NC−14000(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \mathrm{NC}^{-1}
  3. C2002(i^+j^)NC−1200 \sqrt{2}(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \mathrm{NC}^{-1}
  4. D4002(i^+j^)NC−1400 \sqrt{2}(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \mathrm{NC}^{-1}

Correct answer

C. 2002(i^+j^)NC−1200 \sqrt{2}(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \mathrm{NC}^{-1}

Step-by-step solution

E=Kqr2⋅r^=Kqr3⋅rE=\frac{K q}{r^2} \cdot \hat{r}=\frac{K q}{r^3} \cdot r Here, r=(x2−x1)2+(y2−y1)2+(z2−z1)2=(30)2+(30)2+02=302cm=302×10−2m\begin{aligned} r & =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2} \\ & =\sqrt{(30)^2+(30)^2+0^2} \\ & =30 \sqrt{2} \mathrm{ cm} \\ & =30 \sqrt{2} \times 10^{-2} \mathrm{ m} \end{aligned}\text{\text{ and } } q=8 \times 10^{-3} \times 10^{-6} \mathrm{C}Also,Also,\quad \mathbf{r}=(30 \hat{\mathbf{i}}+30 \hat{\mathbf{j}}) \times 10^{-2} \mathrm{ m} So, E=9×109×8×10−3×10−6(302×10−2)3×(30i^+30j^)×10−2=9×8×109×10−1127×22×10−6×103×30(i^+j^)=9×8×102×327×22(i^+j^)=2002(i^+j^)NC−1\begin{aligned} \text{ So, } \mathrm{E} & =\frac{9 \times 10^9 \times 8 \times 10^{-3} \times 10^{-6}}{\left(30 \sqrt{2} \times 10^{-2}\right)^3} \times(30 \hat{\mathbf{i}}+30 \hat{\mathbf{j}}) \times 10^{-2} \\ & =\frac{9 \times 8 \times 10^9 \times 10^{-11}}{27 \times 2 \sqrt{2} \times 10^{-6} \times 10^3} \times 30(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \\ & =\frac{9 \times 8 \times 10^2 \times 3}{27 \times 2 \sqrt{2}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \\ & =200 \sqrt{2}(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \mathrm{NC}^{-1} \end{aligned}$
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