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        When a test charge is placed at the centre of dipole then what will be the net force on the test charge? Please give good explanation.
7 months ago

Ash
36 Points
							Let two charges of equal magnitude Q but opposite sign be placed at 2a from each other.Now a charge q0 is placed midway between them.The force experienced by it due the negative charge is $F=\frac{1}{4\pi \varepsilon}\frac{Qq_{0}}{a^{2}}$ towards the negative chargeThe force experienced by it due the positive charge is $F=\frac{1}{4\pi \varepsilon}\frac{Qq_{0}}{a^{2}}$ away from the postive chargeSo,the net force experienced by it will be$F_{net}=2 \frac{1}{4\pi \varepsilon}\frac{Qq_{0}}{a^{2}}$ towards the negative charge.

3 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions