Question:
In an isosceles right angled triangle ABC, the equal sides AB and AC are 10 cm. Charges \(+5\mu C\), \(+20\mu C\) and \(+20\mu C\) are placed at A, B and C respectively. A charge \(+2\mu C\) is placed at the midpoint of BC. The net electrostatic force on it is:
In an isosceles right angled triangle ABC, the equal sides AB and AC are 10 cm. Charges \(+5\mu C\), \(+20\mu C\) and \(+20\mu C\) are placed at A, B and C respectively. A charge \(+2\mu C\) is placed at the midpoint of BC. The net electrostatic force on it is:
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At the midpoint of the hypotenuse of a right triangle, distances from all three vertices are equal.
Updated On: Jun 18, 2026
- \(24N\)
- \(9N\)
- \(36N\)
- \(18N\)
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The Correct Option is D
Solution and Explanation
Concept:
The midpoint of hypotenuse of a right triangle is equidistant from all three vertices.
Hence,
\[
MA=MB=MC.
\]
The forces due to charges at B and C cancel each other because they are equal and opposite.
Step 1: Find distance from midpoint to vertex.
\[ BC=10\sqrt2. \] \[ MA=\frac{BC}{2} = 5\sqrt2 cm. \] \[ r=5\sqrt2\times10^{-2}m. \]
Step 2: Force due to charge at A.
\[ F= \frac{kq_1q_2}{r^2}. \] \[ = \frac{9\times10^9(5\times10^{-6})(2\times10^{-6})} {(5\sqrt2\times10^{-2})^2}. \] \[ = 18N. \] Hence \[ \boxed{18N}. \]
Step 1: Find distance from midpoint to vertex.
\[ BC=10\sqrt2. \] \[ MA=\frac{BC}{2} = 5\sqrt2 cm. \] \[ r=5\sqrt2\times10^{-2}m. \]
Step 2: Force due to charge at A.
\[ F= \frac{kq_1q_2}{r^2}. \] \[ = \frac{9\times10^9(5\times10^{-6})(2\times10^{-6})} {(5\sqrt2\times10^{-2})^2}. \] \[ = 18N. \] Hence \[ \boxed{18N}. \]
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