COMEDK · Physics · 17. Electrostatics
Point charge \(\sqrt{2}C, \sqrt{2}C\), and \(-2C\) are placed at the three vertices of a right-angled triangle in air. [as shown in the figure below] What is the electric field at a point P on the hypotenuse that is equidistant from all three charges. Given distances \(XP = YP = ZP = 0.5\ m\)

- A \(7\cdot 2 \times 10^9\ NC^{-1}\) along PY
- B \(7\cdot 2 \times 10^{10}\ NC^{-1}\) along PY
- C \(0\cdot 72 \times 10^9\ NC^{-1}\) along YP
- D \(0\cdot 72 \times 10^{10}\ NC^{-1}\) along YP
Answer & Solution
Correct Answer
(B) \(7\cdot 2 \times 10^{10}\ NC^{-1}\) along PY
Step-by-step Solution
Detailed explanation
Let the charges at vertices \(X\), \(Y\), and \(Z\) be \(q_X = \sqrt{2}\ C\), \(q_Y = -2\ C\), and \(q_Z = \sqrt{2}\ C\) respectively. Point \(P\) is on the hypotenuse \(XZ\) and is equidistant from \(X\) and \(Z\), meaning it is the midpoint of \(XZ\). The electric field at…
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