91Ó°ÊÓ

The intensity of the electric field at a point $\mathrm{d}$units away from a point charge $\mathrm{q}$can be calculated using the following formula:

$\left(\mathrm{E}\right)=\mathrm{q}/\left[4{\mathrm{Ï€Î\mu »å}}^2\right]{\mathrm{NC}}^{-1}$

The vector sum of the intensities produced by the individual charges equals the intensity of the electric field at any point due to a number of charges.

Distance between the outer edges is $6.7\mathrm{cm}$.

Diameter of sphere is$2.0\mathrm{cm}$.

Hence the distance between center is$(6.7+2.0)=8.7\mathrm{cm}$.

Mid point is ,

$=(8.7/2)\mathrm{cm}$

localid="1648629844607" $=4.35\mathrm{cm}=0.0435\mathrm{m}$.

Charge of spherelocalid="1648629881080" $1$islocalid="1648391289846" $41\mathrm{nC}$.

Charge of spherelocalid="1648629893533" $2$is$-76\mathrm{nC}$.

The fields will now all point in the same direction.

As a result, whole field is:

localid="1648630029365" ${\mathrm{E}}_{\mathrm{net}}={\mathrm{E}}_1+{\mathrm{E}}_2$

localid="1648630472057" ${\mathrm{E}}_1$due tolocalid="1648630003851" $42\mathrm{nC}$will be:

localid="1648630478720" ${\mathrm{E}}_1=\frac{9×{10}^9×41×{10}^{-9}}{(0.0435{)}^2}$

localid="1648630485528" ${\mathrm{E}}_1=1.95×{10}^5\mathrm{N}/\mathrm{C}$

localid="1648630491962" ${\mathrm{E}}_2$due to localid="1648630497375" $-76\mathrm{nC}$will be:

localid="1648630504401" ${\mathrm{E}}_2=\frac{9×{10}^9×76×{10}^{-9}}{(0.0435{)}^2}$

localid="1648630510652" ${\mathrm{E}}_2=3.61×{10}^5\mathrm{N}/\mathrm{C}$

Substitute${\mathrm{E}}_1$and${\mathrm{E}}_2$values we get,

${\mathrm{E}}_{\text{net}}=1.95×{10}^5\mathrm{N}/\mathrm{C}+3.61×{10}^5\mathrm{N}/\mathrm{C}$

${\mathrm{E}}_{\mathrm{net}}=5.56×{10}^5\mathrm{N}/\mathrm{C}$