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:
The particle comes to rest before reaching the position x=4m.
Explanation:
When the object is at x=0m, all of its energy is kinetic energy. Using the equation for kinetic energy yields KE=1/2mv^2=(12)(2)(3)^2=9J. Using the given equation for potential energy when the object is at x=4m yields U=4x^2=4(4)^2=64J. Since the system only has 9J of energy, the object comes to rest before reaching x=4m.
D= 9.7^3/(69)(4.2)^2
d=912.673/289.8^2
d=912.673/83984.04
d=0.01086721953362...
I hope this helped!
The vertical component is given by the sin of the angle. So this is 20*sin(53)=15.97m/s
