According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.
Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k · q₂/r² = k · (q₁ + q₂) / r²
where:
k = 9×10⁹ N·m²/C²
We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10<span>⁻¹² C
</span><span>r = 5.0 cm = 0.05 m
Now, we can apply the formula:
</span><span>E(r) = k · (q₁ + q₂) / r²
= </span>9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
= 39.6 N/C
Hence, <span>the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C</span>
Answer:
57.1 km/hr
Explanation:
To find the average speed you take the total distance divided by the total elapsed time.
So, the total distance is 140 + 60 = 200
the total elapsed time is found by taking 140/70=2 and 60/40=1.5
2+1.5=3.5
The plug the numbers into the equation,
200/3.5=57.1
Answer:
Just consider the resistance as water flowing through a pipe. If the pipe is too small in radius (consider thin wire as small pipe) water can’t flow easly. If the pipe is big in radius (consider thick wire as big pipe) water can flow easly. So flowing water through a small pipe for a long distance will inversly affect the normal flow where as flowing water through a big pipe for a small distance will not affect too much the normal flow.
So long & thin wire has high resistance. Short & thick wire has low resistance.
B. A wire that is 2 m long and has a cross-sectional area of 0.066
