## anonymous one year ago Two small insulating spheres with radius 5.00×10−2 m are separated by a large center-to-center distance of 0.600 m . One sphere is negatively charged, with net charge -2.10 μC , and the other sphere is positively charged, with net charge 3.70 μC . The charge is uniformly distributed within the volume of each sphere. What is the magnitude E of the electric field midway between the spheres? Take the permittivity of free space to be ϵ0 = 8.85×10−12 C2/(N⋅m2) .

1. Michele_Laino

hint: here is the situation of your problem: |dw:1436365695363:dw|

2. Michele_Laino

where $\begin{gathered} {Q_1} = 3.70 \times {10^{ - 6}}coulombs \hfill \\ {Q_2} = - 2.10 \times {10^{ - 6}}coulombs \hfill \\ \end{gathered}$

3. Michele_Laino

now the total electric field at midpoint M, is given by the subsequent vector sum: $\large {\mathbf{E}}\left( M \right) = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{{{\left( {d/2} \right)}^2}}}{\mathbf{\hat x}} + \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_2}}}{{{{\left( {d/2} \right)}^2}}}{\mathbf{\hat x}}$ |dw:1436367402248:dw|

4. Michele_Laino

oops.. the total electric field at midpoint M is: $\large {\mathbf{E}}\left( M \right) = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{{{\left( {d/2} \right)}^2}}}{\mathbf{\hat x}} - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_2}}}{{{{\left( {d/2} \right)}^2}}}{\mathbf{\hat x}}$

5. Michele_Laino

and the requested magnitude is: $\large E\left( M \right) = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{{{\left( {d/2} \right)}^2}}} - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_2}}}{{{{\left( {d/2} \right)}^2}}}$

6. Michele_Laino

where you have to substitute these quantities: $\Large \begin{gathered} {Q_1} = 3.70 \times {10^{ - 6}} \hfill \\ {Q_2} = - 2.10 \times {10^{ - 6}} \hfill \\ \end{gathered}$

7. anonymous

Yes, it's usually safer to compute each E before adding them. the r = d/2 happens to be the situation here ... it does NOT mean "radius = diameter/2 " <= NO!