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anonymous
 one year ago
Two small insulating spheres with radius 5.00×10−2 m are separated by a large centertocenter 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) .
anonymous
 one year ago
Two small insulating spheres with radius 5.00×10−2 m are separated by a large centertocenter 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) .

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: here is the situation of your problem: dw:1436365695363:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2where \[\begin{gathered} {Q_1} = 3.70 \times {10^{  6}}coulombs \hfill \\ {Q_2} =  2.10 \times {10^{  6}}coulombs \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2oops.. the total electric field at midpoint M is: \[\large {\mathbf{E}}\left( M \right) = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{{{\left( {d/2} \right)}^2}}}{\mathbf{\hat x}}  \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_2}}}{{{{\left( {d/2} \right)}^2}}}{\mathbf{\hat x}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2and the requested magnitude is: \[\large E\left( M \right) = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_1}}}{{{{\left( {d/2} \right)}^2}}}  \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Q_2}}}{{{{\left( {d/2} \right)}^2}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2where you have to substitute these quantities: \[\Large \begin{gathered} {Q_1} = 3.70 \times {10^{  6}} \hfill \\ {Q_2} =  2.10 \times {10^{  6}} \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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!
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