Answer:
66.375 x 10⁻⁶ C/m
Explanation:
Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as
;
∅ = Q / ε₀ -----------------(i)
Where;
∅ = 7.5 x 10⁵ Nm²/C
ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²
Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;
7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)
Solve for Q;
Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²
Q = 66.375 x 10⁻⁷ C
Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e
L = Q / l ----------------------(ii)
Where;
Q = 66.375 x 10⁻⁷ C
l = length of the rod = 10.0cm = 0.1m
Substitute these values into equation (ii) as follows;
L = 66.375 x 10⁻⁷C / 0.1m
L = 66.375 x 10⁻⁶ C/m
Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.
Explanation:
Formula to calculate electric field because of the plate is as follows.
E = 
= 
= 
Now, we will consider that equilibrium of forces are present there. So,
ma = qE
a = 
= 
According to the third equation of motion,
or, d = 
= 
= 0.254 m
Thus, we can conclude that
the proton will travel 0.254 m before reaching its turning point.
The answer would be in the chart or graph A is 1 B is 2
Answer:
<h2>workdone = force × distance </h2><h2>236J = 18.9cos(o) × 24.4</h2><h2>236/24.4 = 18.9cos(o)</h2><h2>(0.5117)cos^-1 = (o)</h2><h2><u>59.21°</u></h2>
