Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE,...
Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.
Problems: 3.7
Point charge \(q\) and \( - q\) are located at the vertices of a square with diagonals \(2l\) as shown in figure. Find the magnitude of the electric field strength at a point located symmetrically with respect to the vertices of the square at a distance \(x\) from its centre.
Solution: 3.7
Let us fix the co-ordinate system by taking the point of intersection of the diagonals as the origin and let \(\vec k\) be directed normally, emerging from the plane of figure.
Hence the sought field strength is
\(\vec E = \frac{q}{{4\pi {\varepsilon _0}}}\frac{{l\hat i + x\hat k}}{{{{\left( {{l^2} + {x^2}} \right)}^{\frac{3}{2}}}}} + \frac{{ - q}}{{4\pi {\varepsilon _0}}}\frac{{l\left( { - \hat i} \right) + x\hat k}}{{{{\left( {{l^2} + {x^2}} \right)}^{\frac{3}{2}}}}} + \frac{{ - q}}{{4\pi {\varepsilon _0}}}\frac{{l\hat j + x\hat k}}{{{{\left( {{l^2} + {x^2}} \right)}^{\frac{3}{2}}}}} + \frac{q}{{4\pi {\varepsilon _0}}}\frac{{l\left( { - \hat j} \right) + x\hat k}}{{{{\left( {{l^2} + {x^2}} \right)}^{\frac{3}{2}}}}}\)
\( = \frac{q}{{4\pi {\varepsilon _0}{{\left( {{l^2} + {x^2}} \right)}^{\frac{3}{2}}}}}\left[ {2l\hat i - 2l\hat j} \right]\)
Thus, \(E = \frac{{ql}}{{\sqrt 2 \pi {\varepsilon _0}{{\left( {{l^2} + {x^2}} \right)}^{\frac{3}{2}}}}}\)
Problems: 3.7
Point charge \(q\) and \( - q\) are located at the vertices of a square with diagonals \(2l\) as shown in figure. Find the magnitude of the electric field strength at a point located symmetrically with respect to the vertices of the square at a distance \(x\) from its centre.
Solution: 3.7
Let us fix the co-ordinate system by taking the point of intersection of the diagonals as the origin and let \(\vec k\) be directed normally, emerging from the plane of figure.
Hence the sought field strength is
\(\vec E = \frac{q}{{4\pi {\varepsilon _0}}}\frac{{l\hat i + x\hat k}}{{{{\left( {{l^2} + {x^2}} \right)}^{\frac{3}{2}}}}} + \frac{{ - q}}{{4\pi {\varepsilon _0}}}\frac{{l\left( { - \hat i} \right) + x\hat k}}{{{{\left( {{l^2} + {x^2}} \right)}^{\frac{3}{2}}}}} + \frac{{ - q}}{{4\pi {\varepsilon _0}}}\frac{{l\hat j + x\hat k}}{{{{\left( {{l^2} + {x^2}} \right)}^{\frac{3}{2}}}}} + \frac{q}{{4\pi {\varepsilon _0}}}\frac{{l\left( { - \hat j} \right) + x\hat k}}{{{{\left( {{l^2} + {x^2}} \right)}^{\frac{3}{2}}}}}\)
\( = \frac{q}{{4\pi {\varepsilon _0}{{\left( {{l^2} + {x^2}} \right)}^{\frac{3}{2}}}}}\left[ {2l\hat i - 2l\hat j} \right]\)
Thus, \(E = \frac{{ql}}{{\sqrt 2 \pi {\varepsilon _0}{{\left( {{l^2} + {x^2}} \right)}^{\frac{3}{2}}}}}\)

COMMENTS