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Equal charges q  are placed at the four corners A,B,C,D  of a square of length a . The magnitude of the force on the charge at B will be

  • Option 1)

    \frac{3q^{2}}{4\pi \varepsilon _{0}a^{2}}\;

  • Option 2)

    \; \frac{4q^{2}}{4\pi \varepsilon _{0}a^{2}}\;

  • Option 3)

    \; \left ( \frac{1+2\sqrt{2}}{2} \right )\frac{q^{2}}{4\pi \varepsilon _{0}a^{2}}\; \;

  • Option 4)

    \; \left ( 2+\frac{1}{\sqrt{2}} \right )\frac{q^{2}}{4\pi \varepsilon _{0}a^{2}}\;

 

Answers (1)

best_answer

As we learned

Magnitude of the Resultant force -

F_{net}=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\cos \Theta }

- wherein

 

 After following the guidelines mentioned above

 

F_{net}=F_{AC}+F_{D}=\sqrt{F_{A}^{2}+F_{C}^{2}}+F_{D}

Since\; \; F_{A}=F_{C}=\frac{kq^{2}}{a^{2}}\; \; and\; \; F_{D}=\frac{kq^{2}}{(a\sqrt{2})^{2}}

F_{net}=\frac{\sqrt{2}kq^{2}}{a^{2}}+\frac{kq^{2}}{2a^{2}}=\frac{kq^{2}}{a^{2}}\left ( \sqrt{2}+\frac{1}{2} \right )=\frac{q^{2}}{4\pi \varepsilon _{0}a^{2}}\left ( \frac{1+2\sqrt{2}}{2} \right )

 


Option 1)

\frac{3q^{2}}{4\pi \varepsilon _{0}a^{2}}\;

Option 2)

\; \frac{4q^{2}}{4\pi \varepsilon _{0}a^{2}}\;

Option 3)

\; \left ( \frac{1+2\sqrt{2}}{2} \right )\frac{q^{2}}{4\pi \varepsilon _{0}a^{2}}\; \;

Option 4)

\; \left ( 2+\frac{1}{\sqrt{2}} \right )\frac{q^{2}}{4\pi \varepsilon _{0}a^{2}}\;

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Aadil

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