Skip to content
Numericals · Q19

Q.A charge +q exerts a force of magnitude -0.2 N on another charge -2q. If they are separated by 25.0 cm, determine the value of q.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
79% · 19/24 Questions
✓ Free question

The magnitude of the force between charge +q+q and charge −2q-2q, separated by r=0.25r=0.25 m, is (taking magnitudes, since the question gives the force's magnitude as 0.2 N, with the negative sign in the problem simply indicating the attractive nature since the charges are of opposite sign) F=14πϵ0(q)(2q)r2=2kq2r2.F=\frac{1}{4\pi\epsilon_0}\frac{(q)(2q)}{r^2}=\frac{2kq^2}{r^2}. Substituting F=0.2F=0.2 N, r=0.25r=0.25 m, k=9×109k=9\times10^9: 0.2=2(9×109)q2(0.25)2=1.8×1010q20.0625=2.88×1011q2  ⇒  q2=0.22.88×1011=6.944×10−13.0.2=\frac{2(9\times10^9)q^2}{(0.25)^2}=\frac{1.8\times10^{10}q^2}{0.0625}=2.88\times10^{11}q^2\;\Rightarrow\;q^2=\frac{0.2}{2.88\times10^{11}}=6.944\times10^{-13}. Taking the square root, q=6.944×10−13≈8.333×10−7 C=0.8333 μC,q=\sqrt{6.944\times10^{-13}}\approx8.333\times10^{-7}\,\text{C}=0.8333\,\mu\text{C}, which matches the source's own printed answer exactly, confirming both the method and the given numbers are consistent here. [!ANSWER] q≈0.8333 μCq\approx0.8333\,\mu C.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.