Answer:
Explanation:
A point charge of +2q centered in a conductive spherical shell of inner diameter a and outer diameter b will induce - 2q charge on the inner surface and +2q charge on the outer surface of the shell. Since 8q charge has been added to the shell , this charge will reside on the outer surface of the shell. so total charge on the outer surface will be 10q. At a point less than a , the electric field will be due to +2q charge situated at the centre . The electric field will be as follows
E = k .2q / r² for r < a
= 8kq/ a²
electric field at a point r = a>b
total charge lying inside is +2q - 2q = 0 . So in the thickness of the shell , electric field will be zero as total charge inside is nil.
For a point at r > b total charge inside is 2q-2q+10q = 10q , so electric field at r which is lying outside the shell .
E = k 10 q / r² for r > b
Answer:
Explanation:
The equation for this is
f = μ where f is the frictional force the block needs to overcome, μ is the coefficient of static friction, and (that means that the normal force is the same as the weight of the block which has an equation of weight = mass times the pull of gravity). Filling in:
1.09 = μ(.413)(9.8) and
μ = so
μ = .27
Answer:
c. No. An equation may have consistent units but still be numerically invaid.
Explanation:
For an equation to be corrected, it should have consistent units and also be numerically correct.
Most equation are of the form;
(Actual quantity) = (dimensionless constant) × (dimensionally correct quantity)
From the above, without the dimensionless constant the equation would be numerically wrong.
For example; Kinetic energy equation.
KE = 0.5(mv^2)
Without the dimensionless constant '0.5' the equation would be dimensionally correct but numerically wrong.