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  1. Problems
  2. /A closed surface encloses charges of +7.0 nC, -1.5 nC and +1.0 nC…
MediumPhysicsHigh School (11–12)Electricity & Magnetism › ElectrostaticsNumericalConceptualNEETBoard ExamAPA-LevelIBAEEE (Amrita)
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A closed surface encloses charges of +7.0 nC, -1.5 nC and +1.0 nC…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A closed surface encloses charges of +7.0+7.0+7.0 nC, −1.5-1.5−1.5 nC and +1.0+1.0+1.0 nC. A fourth charge of +11.0+11.0+11.0 nC is outside the surface. Find the net flux through the surface. The fourth charge is then pushed inside the surface. What is the net flux now? Does the field at a point on the surface change when the fourth charge is moved?

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

N m²/C
N m²/C
Need a reminder? The key ideaShowHide

Gauss's law: the net flux through a closed surface is Φ=qin/ε0\Phi = q_{\text{in}}/\varepsilon_0Φ=qin​/ε0​, with ε0=8.85×10−12 C2/(N m2)\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)ε0​=8.85×10−12 C2/(N m2). Only the net charge inside counts. The shape and size of the surface do not matter. For a sphere of radius rrr with the charge at its centre, E=Φ/(4πr2)E = \Phi/(4\pi r^2)E=Φ/(4πr2).

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. The first case. The charge inside is 7.0−1.5+1.0=6.57.0 - 1.5 + 1.0 = 6.57.0−1.5+1.0=6.5 nC. The charge outside adds no net flux. So Φ=6.5×10−9 C8.85×10−12 C2/(N m2)=734 N m2/C\Phi = \dfrac{6.5 \times 10^{-9}\ \text{C}}{8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)} = 734\ \text{N m}^2/\text{C}Φ=8.85×10−12 C2/(N m2)6.5×10−9 C​=734 N m2/C.

Step 2. The second case. Now the charge inside is 6.5+11.0=17.56.5 + 11.0 = 17.56.5+11.0=17.5 nC. So Φ=17.5×10−98.85×10−12=1977 N m2/C\Phi = \dfrac{17.5 \times 10^{-9}}{8.85 \times 10^{-12}} = 1977\ \text{N m}^2/\text{C}Φ=8.85×10−1217.5×10−9​=1977 N m2/C.

Step 3. The field. Yes, the field at a point on the surface changes. The fourth charge makes a field of its own at every point, and moving the charge changes it. Gauss's law tells us only about the net flux, which is a sum over the whole surface.

Answer: 734 N m2/C734\ \text{N m}^2/\text{C}734 N m2/C and 1977 N m2/C1977\ \text{N m}^2/\text{C}1977 N m2/C. The field at a point of the surface does change.

Check. The flux grew by 11.011.011.0 nC divided by ε0\varepsilon_0ε0​, which is 1243 N m2/C1243\ \text{N m}^2/\text{C}1243 N m2/C. And 734+1243=1977734 + 1243 = 1977734+1243=1977. It fits.

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  1. Gauss's lawElectric Flux and Gauss's Law

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