four charges are placed at the corners of a square of side 'a' . Find the net force on the charge at the corner C. The charge at C is -q and the charge at A,B and D is +q, +q and +q respectively
Answers
Net force = (k q²/d²) ( √2 + 1/2) Direction from C to A
Explanation:
F = k q₁(q₂)/r²
r = Distance between Charges
q₁ , q₂ are charges
d = Distance between them
Repulsive force if same sign between charges
Attractive force if opposite sign between charges
k = 1/4π∈₀ = 9.0 x 10⁹ N·m²/C²
Force by D on C attractive Force
= k q²/d² ( Direction from C to D)
Force by B on C attractive Force
= k q²/d² ( Direction from C to B)
Force by A on C attractive Force
= k q²/(d√2)² ( Direction from C to A)
= k q²/2d² ( Direction from C to A)
Net Force of k q²/d² ( Direction from C to D) & k q²/d² ( Direction from C to B)
= √ (k q²/d²)² + ( k q²/d²)²
=√2 . k q²/d²
net force on the charge at the corner C = k q²/2d² + √2 . k q²/d²
= (k q²/d²) ( √2 + 1/2) Direction from C to A
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Explanation:
F due to B and C = k q²/a²
F due to D and C = k q²/a²
Resultant of F due B and C & D and C is:
= √F²(bc) + F²(dc) + 2F.F.cos90*
= √(k q²/r²)² + (kq²/r²)²
= √2 k q²/r²
F due to A is along diagonal(length = a√2).
F due to A = k q²/(√2a)²
Hence, total force on C is:
= √2 kq²/r² + kq²/2a²
= kq²/r² (√2 + 1/2) ,
Or, k(√2 + 1/2) q²/r² , where k = 1/4πε