Answer:
0.546 ohm / μm
Explanation:
Given that :
N = 1.015 * 10^17
Electron mobility, u = 3900
Hole mobility, h = 1900
Ng = 4.42 x10^22
q = 1.6*10^-19
Resistivity = 1/qNu
Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)
= 0.01578880889 ohm /cm
Resistivity of germanium :
R = 1 / 2q * sqrt(Ng) * sqrt(u*h)
R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)
R = 1 /0.0001831
R = 5461.4964 ohm /cm
5461.4964 / 10000
0.546 ohm / μm
The spring scale will read 559 Newton's or 125.7 pounds.
In order to find the force (F), you would have to use the formula for it:
F=ma
where m is mass and a is acceleration.
In the problem, the mass is 2.85kg and the acceleration is 4.9m/s^2.
Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.
let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.
the mathematical formula for potential is 
for positive charges the potential is positive and is negative for negative charges.
the formula for electric field is given as-
for positive charges,the line filed is away from it and for negative charges the filed is towards it.
we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.
here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.
now let the two charges of same nature.let these are positively charged.
here we can not find a point between two charges and on the line joining two charges where the potential is zero.
but at the mid point of the line joining two charges the filed is zero.