Answer:
Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:
lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:
To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):
=![\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7B4%5Cpi%20e_%7B0%7D%20%7D%20%5B%5Cfrac%7B1%7D%7BR%7D%20-%5Cfrac%7Br%5E%7B2%7D-R%5E%7B2%7D%20%20%7D%7B2R%5E%7B3%7D%20%7D%20%5D)
∴NOTE: Graph is attached
Answer:
2.16×10⁻⁶ N
Explanation:
Applying,
F = kqq'/r² (coulomb's Law)....................... Equation 1
Where F = electrostatic force, k = coulomb's constant, q = charge on the styrofoam, q' = charge on the grain of salt, r = distance between the charges.
From the question,
Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m
Constant: k = 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = (2.0×10⁻⁶)(3.0×10⁻¹¹)(8.99×10⁹)/0.5²
F = 2.16×10⁻⁶ N
Mechanical energy equals the sum of potential and kinetic energy. During the process, all PE converts into KE, assuming air resistance is neglected. So, the mechanical energy does not change and is equal to the initial potential energy.
ME
=mgh
=0.005 x 9.81 x 3
=0.147J
Answer:
R = m⁴/kg . s
Explanation:
In this case, the best way to solve this is working with the units in the expression.
The units of velocity (V) are m/s
The units of density (d) are kg/m³
And R is a constant
If the expression is:
V = R * d
Replacing the units and solving for R we have
m/s = kg/m³ * R
m * m³ / s = kg * R
R = m * m³ / kg . s
<h2>
R = m⁴ / kg . s</h2>
This should be the units of R
Hope this helps
Answer:
In D: 3J
Explanation:
Potential energy: Ep=mgh where m is the mass, h altitude.
In point A: h=20cm=0.2m
Epa=12=0.2×mg. Thus mg=12/0.2=60N
For point D: hd=5cm=0.05m
Epd=mg×0.05=60×0.05=3J
