Electric field:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{k}\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}}$

**Part (a)**

The electric field at q_{a} due to q_{b} is:

$\begin{array}{rcl}{\mathbf{E}}_{\mathbf{a}\mathbf{b}}& \mathbf{=}& \frac{\mathbf{(}\mathbf{9}\mathbf{\times}{\mathbf{10}}^{\mathbf{9}}\mathbf{)}\mathbf{(}\mathbf{10}\mathbf{.}\mathbf{8}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{)}}{{\mathbf{(}\mathbf{21}\mathbf{\times}\mathbf{10}\mathbf{-}\mathbf{2}\mathbf{)}}^{\mathbf{2}}}\end{array}$

E_{ab} = 2.204 × 10^{6} N/C

The electric field at q_{a} due to q_{c} is:

$\begin{array}{rcl}{\mathbf{E}}_{\mathbf{ac}}& \mathbf{=}& \frac{\mathbf{(}\mathbf{9}\mathbf{\times}{\mathbf{10}}^{\mathbf{9}}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{8}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{)}}{{\mathbf{(}\mathbf{21}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{)}}^{\mathbf{2}}}\end{array}$

E_{ac} = 9.80 × 10^{5} N/C

In the figure, the point charges are located at the corners of an equilateral triangle 21 cm on a side.

Part (a) Find the magnitude of the electric field in N/C at the location of q_{a} given that q_{b}=10.8 uC and q_{c}=4.8 uC.

Part (b) Find the direction of the electric field at q_{a}, in degrees above the negative x-axis with origin at q_{a}

Part (c) What is the magnitude of the force in N on q_{a} given that q_{a}=1.9 nC?

Part (d) What is the direction of the force on q_{a}, in degrees above the negative x-axis with origin at q_{a}?

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