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Class 12 › Physics › Electric Charge and Field
CBSE2022Class Class 12 · Physics3 Marks · Short✅ Verified
A particle of mass 10⁻³ kg and charge 5 µC is thrown at a speed of 20 m/s against a uniform electric field of strength 2 × 10⁵ N/C. How much distance will it travel before coming to rest momentarily?
✅ Answer & Solution
<b>Given:</b><br>• Mass m = 10⁻³ kg<br>• Charge q = 5 µC = 5 × 10⁻⁶ C<br>• Initial speed u = 20 m/s<br>• Electric field E = 2 × 10⁵ N/C<br>• Final speed v = 0 (momentarily at rest)<br><br><b>Step 1: Calculate force on the particle</b><br>F = qE<br>F = (5 × 10⁻⁶) × (2 × 10⁵)<br><b>F = 1 N</b><br><br>(The particle is moving <b>against</b> the field, so force opposes motion → deceleration)<br><br><b>Step 2: Calculate deceleration</b><br>a = F/m<br>a = 1 / 10⁻³<br><b>a = 1000 m/s²</b> (deceleration)<br><br><b>Step 3: Apply third equation of motion</b><br>v² = u² - 2as<br><br><b>Step 4: Substitute values</b><br>0 = (20)² - 2 × 1000 × s<br>0 = 400 - 2000s<br>2000s = 400<br><b>s = 0.2 m</b><br><br><b>Final Answer:</b><br>Distance travelled before momentarily coming to rest = <b>0.2 m = 20 cm</b>