A point charge q1 = -8.9 (mu)C is located at the center of a thick conducting shell of inner...

Question:

A point charge q1 = -8.9 {eq}\mu {/eq}C is located at the center of a thick conducting shell of inner radius a = 2.8 cm and outer radius b = 4.1 cm, The conducting shell has a net charge of q2 = 2.2 {eq}\mu {/eq}C.

1) What is Ex(P), the value of the x-component of the electric field at point P, located a distance 8.5 cm along the x-axis from q1?

2) What is Ey(P), the value of the y-component of the electric field at point P, located a distance 8.5 cm along the x-axis from q1?

3) What is Ey(R), the value of the y-component of the electric field at point R, located a distance 1.4 cm along the y-axis from q1

Electric field

The electric field is the physical vector field through which electrical interactions are transmitted. This is defined as the force per unit of charge that must be carried out in order to displace any electrical charge. In the case of point loads, or spherical load distributions:

{eq}\displaystyle \vec{E}=\frac{kq}{r^2}\hat{r} {/eq}

where {eq}k {/eq} is the Coulomb constant, {eq}q {/eq} is the electric charge that is generating the field, {eq}r {/eq} is the distance to any point in space, and {eq}\hat{r} {/eq} is a unit vector in the direction of the field.

1) and 2) The field associated with a point charge, or spherical distribution of charge, has a radial character, therefore, at the point (8.5; 0) cm the vertical component of the field will be zero and will only have a horizontal component.

{eq}\vec{E}(P)=E_x(P)\hat{i}+E_y(P)\hat{j}\\ \displaystyle E_x(P)=\frac{kq_{net}}{r^2}\\ \displaystyle E_x(P)=\frac{k(q_1+q_2)}{r^2}\\ \displaystyle E_x(P)=\mathrm{\frac{(8.99\cdot10^9\,Nm^2/C^2)(-8.9\cdot10^{-6}\,C+2.2\cdot10^{-6}\,C)}{(8.5\cdot10^{-2}\,m)^2}}\\ \therefore\boxed{E_x(P)=-8.3\cdot10^6\,\mathrm{N/C}}\\ \therefore\boxed{E_y(P)=0\,\mathrm{N/C}}\\ {/eq}

3) The field associated with a point charge, or spherical distribution of charge, has a radial character, therefore, at the point (0; 1.4) cm the horizontal component of the field will be zero and will only have a vertical component.

{eq}\vec{E}(R)=E_x(R)\hat{i}+E_y(R)\hat{j}\\ \displaystyle E_y(R)=\frac{kq_{net}}{r^2}\\ \displaystyle E_y(R)=\frac{k(q_1+q_2)}{r^2}\\ \displaystyle E_y(R)=\mathrm{\frac{(8.99\cdot10^9\,Nm^2/C^2)(-8.9\cdot10^{-6}\,C+2.2\cdot10^{-6}\,C)}{(1.4\cdot10^{-2}\,m)^2}}\\ \therefore\boxed{E_y(R)=-3.1\cdot10^8\,\mathrm{N/C}}\\ \therefore\boxed{E_x(R)=0\,\mathrm{N/C}}\\ {/eq}