NEET-XII-Physics
29: Electric Field and Potential
- #6Two charges 2.0 × 10-6 C and 1.0 × 10-6 C are placed at a separation of 10 cm. Where should a third charge be placed, such that it experiences no net force due to these charges?Ans : Given:
`` {q}_{1}=2.0\times {10}^{-6}\,\mathrm{\,C\,}``
`` {q}_{2}=1.0\times {10}^{-6}\,\mathrm{\,C\,}``
Let the third charge, q, be placed at a distance of x cm from charge q1, as shown in the figure.
By Coulomb's Law, force,
`` F=\frac{1}{4\,\mathrm{\,\pi \,}{\epsilon }_{0}}\frac{{Q}_{1}{Q}_{2}}{{r}^{2}}``
Force on charge q due to q1,
`` F=\frac{9\times {10}^{9}\times 2.0\times {10}^{-6}\times q}{{x}^{2}}``
Force on charge q due to q2,
`` F\text{'}=\frac{9\times {10}^{9}\times {10}^{-6}\times q}{{\left(10-x\right)}^{2}}``
`` ``
According to the question,
`` F-F\text{'}=0``
`` \Rightarrow F=F\text{'}``
`` \Rightarrow \frac{9\times {10}^{9}\times 2\times {10}^{-6}\times q}{{x}^{2}}=\frac{9\times {10}^{9}\times {10}^{-6}\times q}{{\left(10-x\right)}^{2}}``
`` \Rightarrow {x}^{2}=2{\left(10-x\right)}^{2}``
`` \Rightarrow {x}^{2}-40x+200=0``
`` \Rightarrow x=20\pm 10\sqrt{2}``
`` \Rightarrow x=5.9\,\mathrm{\,cm\,}(\because x\ne 20+10\sqrt{2})``
So, the third charge should be placed at a distance of 5.9 cm from q1.
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