## Given ${q}_{1} = 2 {q}_{2} = 3 {q}_{3} = 4 n C$, and each charge is $1 c m$ away from the origin, what is the electric field magnitude at the origin?

Mar 7, 2018

${E}_{o} = \left(2397.33\right) \setminus \hat{x} \setminus \text{V/m" +(1798)\haty \ "V/m}$

#### Explanation:

Electric Field: $E = k \frac{q}{{r}^{2}}$

We simply sum the fields of the three charges at the origin.

Charge q1: ${E}_{1} = k \frac{{q}_{1}}{r} ^ 2$
Charge q2: ${E}_{2} = k \frac{{q}_{2}}{r} ^ 2$
Charge q3: ${E}_{3} = k \frac{{q}_{3}}{r} ^ 2$

The field at the origin: ${E}_{o} = {E}_{1} + {E}_{2} + {E}_{3}$
when ${r}_{1} = {r}_{2} = {r}_{3} = 1$ cm and ${q}_{1} = 2 {q}_{2} = 3 {q}_{3} = 4$ nC.

${E}_{o} = k \left(\left(\frac{4 \times {10}^{- 9}}{1 \times {10}^{- 2}} - \frac{\frac{4}{3} \times {10}^{- 9}}{1 \times {10}^{- 2}}\right) \setminus \hat{x} + \frac{2 \times {10}^{- 9}}{1 \times {10}^{- 2}} \setminus \hat{y}\right)$

${E}_{o} = \left(8.99 \times {10}^{9}\right) \left(\left(\frac{4 \times {10}^{- 9}}{1 \times {10}^{- 2}} - \frac{\frac{4}{3} \times {10}^{- 9}}{1 \times {10}^{- 2}}\right) \setminus \hat{x} + \frac{2 \times {10}^{- 9}}{1 \times {10}^{- 2}} \setminus \hat{y}\right)$

${E}_{o} = \left(8.99 \times {10}^{9}\right) \left(\left(\left(4 \times {10}^{- 7}\right) - \left(\frac{4}{3} \times {10}^{- 7}\right)\right) \setminus \hat{x} + \left(2 \times {10}^{- 7}\right) \setminus \hat{y}\right)$

${E}_{o} = \left(8.99 \times {10}^{9}\right) \left(\left(\frac{8}{3} \times {10}^{- 7}\right) \setminus \hat{x} + \left(2 \times {10}^{- 7}\right) \setminus \hat{y}\right)$

${E}_{o} = \left(2397.33\right) \setminus \hat{x} \setminus \text{V/m" + (1798)\haty \ "V/m}$