A metal sphere of radius 0.1 metre is charged to a potential 36 kilo v...
Given r=0.1m
V=36kv
AB=0.4cm,BC=0.5cm and angle b=90
q=3×10^-6C
To find
V=KQ÷E
Va=KQ÷0.4
Vb=KQ÷0.5
(Va-Vb)q=W
(KQ÷0.4-KQ÷0.5)×2×10^-6
For charged sphere KQ÷0.1=36×10^3
KQ=3600
From 1 and 2
W=3600(0.1÷0.4×0.5)×10^-6×10^2
W=5400×10^-6 =5.4×10^-3J
A metal sphere of radius 0.1 metre is charged to a potential 36 kilo v...
Introduction
To calculate the work required to charge 3 microcoulombs at the corner of triangle ABC, we need to evaluate the potential energy involved in moving the charge in the electric field created by the charged metal sphere.
Given Data
- Radius of the sphere: 0.1 m
- Potential of the sphere: 36 kV (or 36,000 V)
- Charge to be moved: 3 μC (or 3 x 10^-6 C)
- Triangle dimensions:
- AB = 0.4 cm = 0.004 m
- BC = 0.5 m
- Angle B = 90°
Electric Potential
- The potential at a distance 'r' from a charged sphere is given by the formula:
V = k * Q / r
Where k is Coulomb's constant (approximately 8.99 x 10^9 Nm²/C²) and Q is the charge on the sphere.
Calculating Work Done
- Work done (W) in moving a charge in an electric field is given by:
W = q * V
where 'q' is the charge being moved, and 'V' is the potential at the location of the charge.
- In this case:
- V = 36,000 V
- q = 3 x 10^-6 C
- Substitute the values:
W = 3 x 10^-6 C * 36,000 V = 0.108 J
Conclusion
The work required to charge 3 microcoulombs to the corner of triangle ABC is 0.108 Joules.
This calculation assumes that the potential remains constant during the movement of the charge, which is a reasonable approximation given the distances involved relative to the size of the sphere.