A block of mass m having charge q is sliding down an inclined plane of inclination and coefficient of friction.When uniform electric field E is applied parallel to the base of inclined plane,the block slides down with constant velocity.The value of E is?
one would have to have an E force which opposed the weight to just enough to make friction eactly the same as the weight down the plane.
weight down the plane mgsinTheta
force of friction= mg*mu*cosTheta
Now E must reduce weight enough to reduce friction, so
force friction=mgsinTheta
(mg-Eq)*mu=mgsinTheta
E= mg(1-Mu*sinTheta)/q*mu
check that.
I assume you know mu, sinTheta,
To find the value of the electric field E, we can use the concept of force balance.
The forces acting on the block along the inclined plane are:
1. The gravitational force (mg) acting vertically downwards.
2. The normal force (N) exerted by the inclined plane perpendicular to the plane.
3. The friction force (f) opposing the motion of the block.
Since the block slides down with constant velocity, there is no acceleration in the vertical direction. Therefore, the sum of the vertical forces is zero:
N - mgcosθ = 0 ... (Equation 1)
(where θ is the angle of inclination of the plane)
The friction force is given by:
f = μN
(where μ is the coefficient of friction)
When the uniform electric field E is applied parallel to the base of the inclined plane, an additional electric force (Fe) acts on the block:
Fe = qE
(where q is the charge of the block)
Since the block is sliding down with constant velocity, the sum of the forces along the inclined plane is also zero:
mgsinθ - f - Fe = 0 ... (Equation 2)
Combining Equation 1 and Equation 2, we can solve for E:
mgsinθ - μN - qE = 0 (from Equation 2)
mgsinθ - μ(mgcosθ) - qE = 0 (substituting N = mgcosθ from Equation 1)
Simplifying the equation:
mgsinθ - μmgcosθ - qE = 0
E = (mgsinθ - μmgcosθ) / q
Therefore, the value of E is given by (mgsinθ - μmgcosθ) / q.