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Charges -4Q and +9Q are fixed 2 m apart. On the line joining them, the net electric field is zero at a distance x (in metres) from the -4Q charge. Find x.

1. 4 m
2. 2 m
3. 0.8 m
4. 1.2 m

Correct answer: 4 m

Solution

For opposite-sign charges the zero-field point is outside the pair, nearer the weaker charge (-4Q). Taking x from -4Q and (x+2) from +9Q, equate magnitudes: 4/x² = 9/(x+2)². Taking roots gives 2/x = 3/(x+2) -> x = 4 m.

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