Electric Charges Fields And Gausss Law All subtopics

Equation used: $${F}_{E}=\frac{k|{q}_{1}||{q}_{2}|}{{r}^{2}}$$ Given: $${q}_{1}=5.0\times {10}^{-6}\text{ C}$$ $${q}_{2}=8.0\times {10}^{-6}\text{ C}$$ $$r=0.30\text{ m}$$ Working: $${F}_{E}=\frac{(8.99\times {10}^{9})(5.0\times {10}^{-6})(8.0\times {10}^{-6})}{(0.30{)}^{2}}$$ $${F}_{E}=\frac{3.596\times {10}^{-1}}{0.09}$$ $${F}_{E}=4.00\text{ N}$$ The charges have opposite signs, so the force is attractive. The force on ${q}_{1}$ is directed toward ${q}_{2}$.

The electrostatic force is inversely proportional to ${r}^{2}$. Tripling the distance means $r$ is replaced by $3r$. So the force becomes $\frac{1}{(3{)}^{2}}=\frac{1}{9}$ of the original force. The force decreases to one-ninth its original value.

The claim is incorrect. Both protons carry positive charge (+e). Two charges of the same sign exert repulsive forces on each other. The electrostatic force between two protons is always repulsive, regardless of separation distance.

Starting point: Coulomb’s law and vector superposition. $$F=\frac{k{q}^{2}}{{d}^{2}}$$ Step 1 — Find the magnitude of each individual force Each of the other two charges is a distance $d$ away and has the same charge $q$. $${F}_{1}={F}_{2}=\frac{k{q}^{2}}{{d}^{2}}$$ Step 2 — Find the angle between the two force vectors In an equilateral triangle, the angle at each vertex is 60°. Both forces are repulsive and directed away from the other charges. The angle between the two force vectors is 60°. Step 3 — Add the two vectors using the law of cosines The angle between the two vectors when placed tail-to-tail is 60°. The resultant magnitude is: $${F}_{\text{net}}=\sqrt{{F}_{1}^{2}+{F}_{2}^{2}+2{F}_{1}{F}_{2}cos60^{\circ}}$$ $${F}_{\text{net}}=\sqrt{{\left(\frac{k{q}^{2}}{{d}^{2}}\right)}^{2}+{\left(\frac{k{q}^{2}}{{d}^{2}}\right)}^{2}+2{\left(\frac{k{q}^{2}}{{d}^{2}}\right)}^{2}\left(\frac{1}{2}\right)}$$ $${F}_{\text{net}}=\frac{k{q}^{2}}{{d}^{2}}\sqrt{1+1+1}$$ $${F}_{\text{net}}=\frac{\sqrt{3} k{q}^{2}}{{d}^{2}}$$ The net force on any vertex charge points directly away from the center of the triangle, by symmetry.

The +Q at the midpoint is a distance $d/2$ from both other charges. The +Q charge repels it — pushing it toward the −Q side. The −Q charge attracts it — also pulling it toward the −Q side. Both forces point in the same direction. The net force on the third charge is directed from +Q toward −Q along the line of separation.

Starting point: Coulomb’s law and Newton’s law of gravitation. $${F}_{E}=\frac{k{e}^{2}}{{d}^{2}}$$ $${F}_{g}=\frac{G{m}_{e}^{2}}{{d}^{2}}$$ Step 1 — Form the ratio $$\frac{{F}_{E}}{{F}_{g}}=\frac{\frac{k{e}^{2}}{{d}^{2}}}{\frac{G{m}_{e}^{2}}{{d}^{2}}}$$ Step 2 — Simplify The ${d}^{2}$ terms cancel: $$\frac{{F}_{E}}{{F}_{g}}=\frac{k{e}^{2}}{G{m}_{e}^{2}}$$ The ratio depends only on $k$, $e$, $G$, and ${m}_{e}$. It is independent of the separation distance.

Gravity is not fundamentally stronger. The electrostatic force between individual charged particles is enormously larger than their gravitational attraction. Gravity dominates at astronomical scales because large objects tend to be electrically neutral. Positive and negative charges cancel, so no net electrostatic force acts between them. Mass is always positive and cannot cancel, so gravitational forces accumulate with increasing mass.

The external field exerts a force on the electrons in each atom, shifting them slightly in the direction opposite to the field. This creates a slight separation of positive and negative charge within each atom - this is polarization. However, in an insulator, electrons cannot move freely from atom to atom. No charge enters or leaves the sphere. The sphere remains electrically neutral overall.

The electrostatic force decreases. Coulomb’s law uses ${\epsilon }_{0}$ in the denominator for vacuum: $F=\frac{|{q}_{1}||{q}_{2}|}{4\pi {\epsilon }_{0}{r}^{2}}$. When a material with $\epsilon >{\epsilon }_{0}$ fills the space, the polarization of the material partially cancels the field between the charges. Replacing ${\epsilon }_{0}$ with a larger $\epsilon $ increases the denominator, which decreases the force.

The result is consistent. The ratio ${F}_{E}/{F}_{g}=k{e}^{2}/(G{m}_{e}^{2})$ is constant for a given pair of particles. It does not depend on distance because both forces share the same $1/{r}^{2}$ dependence. This means for any two charged particles, ${F}_{E}$ always dominates ${F}_{g}$ at every distance. Gravity dominates at astronomical scales not because of distance, but because large objects are electrically neutral. When net charge is zero, ${F}_{E}=0$ and only ${F}_{g}$ remains. The distance-independence of the ratio is fully consistent with this explanation.

Total charge before contact $${Q}_{\text{total}}=+6 \mu \text{C}+(-2 \mu \text{C})=+4 \mu \text{C}$$ Identical conductors share charge equally $${Q}_{\text{each}}=\frac{+4 \mu \text{C}}{2}=+2 \mu \text{C}$$ Each sphere carries +2 μC after separation.

Equal to 8 nC in magnitude. Conservation of charge requires that every electron removed from the glass transfers to the silk. The silk gains −8 nC, so its magnitude equals 8 nC.

The positive rod attracts free electrons in the metal sphere toward the near side. The near side becomes negatively charged; the far side becomes positively charged. The net charge on the sphere remains zero. However, the negative charges on the near side are closer to the rod than the positive charges on the far side. Because electrostatic force is stronger at shorter distances, the attractive force on the near side exceeds the repulsive force on the far side. The sphere experiences a net attractive force toward the rod.

Yes, an insulator can be polarized. Although electrons in an insulator cannot move freely between atoms, the electron cloud around each molecule shifts slightly. The side of each molecule nearest the external charge develops a small induced charge of opposite sign. Across many molecules, this produces a net surface charge and a weak attraction.

The sphere lost 3 μC of negative charge. Because charging involves electron transfer, electrons moved from the sphere to its surroundings. The number of electrons transferred $$n=\frac{\Delta q}{e}=\frac{3\times {10}^{-6}}{1.60\times {10}^{-19}}$$ $$n\approx 1.88\times {10}^{13}\text{ electrons}$$ Approximately 1.88 × 10¹³ electrons left the sphere.

The total charge cannot change. Conservation of charge requires that net charge remains constant when no charge crosses the system boundary. The total remains +3 μC regardless of any internal redistribution.

Electrons flow from the sphere through the wire into Earth. Earth is so large that the added electrons produce no measurable change in Earth’s charge. The sphere’s excess negative charge is neutralized. The sphere’s final charge is zero.

The sphere ends up neutral instead of negatively charged. While the rod was nearby, electrons flowed from Earth into the sphere. When the rod is removed first, the external field disappears and nothing holds the extra electrons on the sphere. Those electrons flow back to Earth through the still-connected ground wire. The sphere returns to its original neutral state.

Yes. At every stage, the total charge of the rod-sphere-Earth system remains constant. Electrons flow between the sphere and Earth, but no charge is created or destroyed. The final state — neutral sphere, unchanged rod, unchanged Earth — has the same total charge as the initial state.

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.