Answer:
B. an action-reaction force pair
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:
a) 3.210J/K
b) -3.073J/K
c) -3.131J/K
d) O
Explanation:
The relationship between entropy and the energy transferred is applied.
Mathematically, dS = dQ/T
the steps are as shown in the attachment.
Answer:
1.13 mA
Explanation:
Length of wire L = 20.5 cm = 0.205m
Radius of wire r = 2.60/2 = 1.3cm = 0.0130m
Voltage V = 1 × 10³ V
Resistivity of pure silicon p = 2300 Ohms • m
Cross sectional area of the wire
A = pi × r² = pi × (0.013)² = 5.307 × 10 ^-4 m²
Resistance of the material
R = p• L/A
= 2300 • 0.205/5.307 × 10^-4 = 0.888 × 10⁶ Ohms
Using Ohms Law
R = V/ I
I = V/R
I = 10³/0.888 × 10⁶
= 0.001126 A
= 1.13 mA