Respuesta :
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 = 3.0×10⁻¹² C
r = 5.0 cm = 0.05 m
Now, we can apply the formula:
E(r) = k · (q₁ + q₂) / r²
= 9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
= 39.6 N/C
Hence, the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C
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 = 3.0×10⁻¹² C
r = 5.0 cm = 0.05 m
Now, we can apply the formula:
E(r) = k · (q₁ + q₂) / r²
= 9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
= 39.6 N/C
Hence, the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C
The magnitude of the electric field 5.0 cm from the center of the two surfaces is 39.6 N/C.
Given :
[tex]\rm q_1 = 8 \; pC[/tex]
[tex]\rm q_2 = 3\;pC[/tex]
[tex]\rm r = 5\;cm[/tex]
Solution :
When you are outside the two spheres, the electric field will be the overlapping of the two electric fields that is,
[tex]\rm E = k\times \dfrac{q_1}{r^2} + k \times \dfrac{q_2}{r^2}[/tex]
[tex]\rm E = \dfrac{k}{r^2} \times (q_1+q_2)[/tex] -------- (1)
To find the magnitude of the electric field 5 cm from the center of the two surfaces we have,
[tex]\rm K = 9 \times 10^9\; Nm^2/C^2[/tex]
[tex]\rm q_1 = 8 \; pC = 8 \times 10^-^1^2\; C[/tex]
[tex]\rm q_2 = 3\;pC = 3\times 10^-^1^2\;C[/tex]
[tex]\rm r = 5\;cm= 0.05\; m[/tex]
Now putting the above values in equation (1) we get,
[tex]\rm E = 9 \times 10^9 \times \dfrac{((8\times 10^-^1^2)+(3\times 10^-^1^2))}{(0.05)^2}[/tex]
[tex]\rm E = 39.6\;N/C[/tex]
Therefore, the magnitude of the electric field 5.0 cm from the center of the two surfaces is 39.6 N/C.
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https://brainly.com/question/8971780?referrer=searchResults
