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True [87]
3 years ago
15

Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k =

8.99 × 1012N/C (figure is in attached file)
Physics
1 answer:
Pavlova-9 [17]3 years ago
4 0

Answer:

The magnitude of the electric field is 8.99*10^{12}N/C in the r direction.

Explanation:

The equation of the electric field is given by:

|\vec{E}|=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant is 8.99 *10^{9}\: Nm^{2}C^{-2}
  • q is the charge
  • r is the distance from A to q

|\vec{E}|=8.99*10^{9}\frac{12.5}{0.11^{2}}    

\vec{E}=9.29*10^{12} \vac{r} \: N/C

Therefore, the magnitude of the electric field is 8.99*10^{12}N/C in the r direction.

I hope it helps you!

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