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>
W=F*s
W=9000*0
W=0
W:Work
F:Force
s:Distance
If you are working with electricity most people will measure it in watts.
So, A. watts is your best answer.
Hope I helped! ^w^
Newton's first law can be taken to mean that if something is moving it tends to keep moving. if at rest it tends to stay at rest.
so, in a car, you and the car are both moving, say at constant speed. Now you're not actually connected to the car as in clamped to it, not yet at least. You're simply sitting in it at rest with respect to it.
but, someone slams on the brakes for whatever reason. The car slows down/stops. what do you do ? well, you would keep going. and moving a few feet in a car can be dangerous, esp if you're moving at high speed. Unless of course you're clamped to the seat, and the seat is clamped to the car and the car is clamped together. then when the car brakes, yes you'll feel the braking effect, but the belt will restrict your movement, keeping you safe, if shocked and bruised.