Electric Fields

Equation used: $$E=\frac{{F}_{E}}{q}$$ Given: $${F}_{E}=4.0\times {10}^{-3}\text{ N}$$ $$q=2.0\times {10}^{-6}\text{ C}$$ Working: $$E=\frac{4.0\times {10}^{-3}}{2.0\times {10}^{-6}}$$ $$E=2.00\times {10}^{3}\text{ N/C}$$ The electric field is 2.00 × 10³ N/C, directed east. The field points east because the test charge is positive. The force on a positive test charge is in the same direction as the field.

The force is to the left. The force on a charge in an electric field is ${\vec{F}}_{E}=q\vec{E}$. A negative value of $q$ reverses the direction. So the force on a negative charge is opposite to the field direction.

The net electric field at the origin is zero. Each positive charge produces a field pointing away from itself. At the origin, the field from the left charge points to the right. The field from the right charge points to the left. These fields are equal in magnitude because the charges and distances are equal. The vector sum is zero.

Field lines cannot cross. The electric field has exactly one direction at every point. If two lines crossed, the field would have two directions at the crossing point. That is physically impossible. So field lines never intersect.

Starting point: For a spherically symmetric charge distribution, the field outside equals that of a point charge at the center (by the shell theorem / Gauss’s law). Step 1 — Apply Coulomb’s law for a point charge $Q$ at distance $r$ The field from a point charge is: $$E=\frac{kQ}{{r}^{2}}$$ Step 2 — State the condition This applies for $r\ge R$, where all charge $Q$ resides on the surface. Result: $$E=\frac{kQ}{{r}^{2}} \text{for }r\ge R$$ The field outside the sphere is identical to that of a point charge $Q$ located at the sphere’s center.

The field is perpendicular to the surface. If a component existed parallel to the surface, free charges in the conductor would experience a net force along the surface. They would move until that parallel component vanished. In electrostatic equilibrium, no charge moves. So only the perpendicular component remains.

The field is directly proportional to $Q$ because $E=kQ/{r}^{2}$. Doubling the charge doubles the field at every point outside the sphere. At $r=3R$, the field is exactly twice its original value.

The field at the center is zero. By symmetry, every small element of charge has a corresponding element on the opposite side of the center. Their field contributions at the center are equal in magnitude and opposite in direction. The vector sum cancels to zero.

The field inside the insulator can be nonzero, while the field inside the conductor is always zero. In a conductor, free charges redistribute until internal forces vanish. In an insulator, charges are fixed and cannot redistribute. Fixed internal charges produce fields that are not canceled.

The field at $r=2R$ is the same for both spheres. Outside a spherically symmetric charge distribution, the field depends only on the total enclosed charge and distance from the center. Both spheres have total charge $+Q$ and spherical symmetry. At $r=2R$, both produce $E=kQ/(2R{)}^{2}=kQ/(4{R}^{2})$. This is consistent with the expression $E=kQ/{r}^{2}$ because that result applies to any spherically symmetric distribution at points outside the charge.