Electrostatics MCQs with Answers – Part 1 (NEET Physics)

Explanation: A negatively charged body has an excess of electrons compared with the number required for electrical neutrality. During ordinary charging by rubbing, electrons can transfer from one material to another. Protons remain bound within atomic nuclei and do not move from one body to another in this process. Since the strip was initially neutral but becomes negative, it must have gained electrons from the cloth. The cloth consequently loses the same number of electrons and becomes positively charged. If the strip acquires a charge \(-Q\), the cloth acquires \(+Q\), provided the two bodies form an isolated system and no charge escapes. The amount of charge transferred is equal in magnitude on the two bodies, although their masses and dimensions may differ. Rubbing facilitates charge transfer; it does not create electric charge. The important distinction is between redistribution of existing electrons and creation of new positive or negative charge.

Explanation: Electrostatic repulsion is a more conclusive test of charge sign than attraction. A positively charged rod repels another body only when that body also has positive net charge under the stated electrostatic conditions. Therefore, the observation establishes that \(P\) is positively charged. Attraction, however, does not uniquely establish the net charge of the attracted body. The positively charged rod can attract a negatively charged body because unlike charges attract. It can also attract an electrically neutral body by inducing a redistribution of charge within that body. In a neutral body, negative charge is displaced slightly towards the rod, while positive charge is effectively displaced farther away. The attraction of the nearer negative charge can exceed the repulsion of the more distant positive charge. This gives a net attractive force even though the body's total charge remains zero. Consequently, the attraction of \(Q\) permits both negative and neutral initial charge conditions. The observation cannot distinguish these two possibilities without an additional test.

Explanation: \(\textbf{Key conditions:}\) Two independent physical requirements must be satisfied: quantisation of charge and conservation of total charge. Since both bodies are initially neutral, their combined initial charge is \[ Q_i=0+0=0 \] Because the two-body system is isolated, \[ Q_f=q_1+q_2=0 \] \(\textbf{Quantisation check:}\) The net charge on each body must be an integral multiple of elementary charge: \[ q=ne,\qquad n=0,\pm1,\pm2,\ldots \] For option A, \[ \frac{3.2\times10^{-19}}{1.6\times10^{-19}}=2 \] Thus the final charges correspond to \(+2e\) and \(-2e\). Their total is \[ Q_f=(+2e)+(-2e)=0 \] Both necessary conditions are satisfied. In option B, each charge has magnitude \(1.5e\), which violates quantisation for an individual isolated body. Option C contains integral charges but has nonzero total charge, whereas option D also has nonzero total charge. \(\textbf{Final answer:}\) The allowed pair is \((+3.2,-3.2)\times10^{-19}\,\text{C}\), corresponding to transfer of two electrons from the first body to the second. The important distinction is that a pair of charges may satisfy conservation without satisfying quantisation, or satisfy quantisation without satisfying conservation.

Explanation: \(\textbf{Initial charges:}\) \[ q_A=+8e,\qquad q_B=-5e,\qquad q_C=-6e \] The initial total charge is obtained by algebraic addition: \[ Q_i=q_A+q_B+q_C \] \[ Q_i=8e-5e-6e=-3e \] \(\textbf{Electron-transfer direction:}\) Five electrons move from \(B\) to \(A\). An electron carries charge \(-e\), so gaining five electrons changes the charge of \(A\) by \(-5e\). Hence, \[ q_A'=+8e-5e=+3e \] Body \(B\) loses five electrons and therefore becomes more positive: \[ q_B'=-5e+5e=0 \] Body \(C\) does not participate in the transfer: \[ q_C'=-6e \] \(\textbf{Conservation check:}\) \[ Q_f=q_A'+q_B'+q_C' \] \[ Q_f=3e+0-6e=-3e=Q_i \] The change in charge on one body is exactly compensated by the opposite change on the other body. \(\textbf{Final answer:}\) \[ \boxed{(q_A',Q_f)=(+3e,-3e)} \]

Explanation: Statement I is correct because conservation of charge applies to the algebraic sum of the charges of all bodies in an isolated system. When electrons move from one body to another, the charge gained by one body is equal and opposite to the charge lost by the other. An individual body may become neutral without violating this principle. The decisive requirement is that the total charge before and after the interaction remains the same. Statement II is incorrect because conservation of electric charge does not imply conservation of the number of charged bodies. For example, two identical conducting spheres initially carrying equal and opposite charges can both become neutral after being brought into contact and separated. Initially, both spheres are charged, whereas finally neither has any net charge. Conversely, a charged conducting sphere can share its charge with an identical neutral sphere, leaving both spheres charged. These processes change the number of charged bodies without changing the total electric charge. Therefore, Statement I is correct and Statement II is incorrect. Charge conservation concerns the total algebraic quantity, not the number of objects carrying nonzero charge.

Explanation: \(\textbf{Initial information:}\) \[ q_A=+11\,\mu\text{C},\qquad q_B=-5\,\mu\text{C} \] The spheres are identical conductors, so after contact and electrostatic equilibrium they have equal charges. \(\textbf{Charge conservation:}\) \[ Q=q_A+q_B \] \[ Q=11-5=+6\,\mu\text{C} \] The final charge on each sphere is \[ q_A'=q_B'=\frac{Q}{2} \] \[ q_A'=q_B'=\frac{6}{2}=+3\,\mu\text{C} \] \(\textbf{Determine electron movement:}\) Sphere \(A\) changes from \(+11\,\mu\text{C}\) to \(+3\,\mu\text{C}\). Its charge change is \[ \Delta q_A=3-11=-8\,\mu\text{C} \] The negative change means that sphere \(A\) gains electrons. Sphere \(B\) must therefore lose the same number of electrons. \(\textbf{Calculate electron number:}\) \[ n=\frac{|\Delta q_A|}{e} \] \[ n=\frac{8\times10^{-6}}{1.6\times10^{-19}} \] \[ n=5.0\times10^{13} \] \(\textbf{Physical check:}\) Electron movement must occur from the initially negative sphere towards the initially highly positive sphere. The final positive charges on both spheres are consistent with the conserved total charge of \(+6\,\mu\text{C}\). \(\textbf{Final answer:}\) \[ n=5.0\times10^{13}\ \text{electrons} \] The electrons move from sphere \(B\) to sphere \(A\).

Explanation: The initial sphere contains equal amounts of positive and negative charge and is electrically neutral. When the negatively charged rod approaches, free electrons in the sphere are repelled towards the side farther from the rod. The side nearer the rod consequently develops induced positive charge, while the farther side develops induced negative charge. This charge separation alone does not change the net charge of the sphere. In Experiment I, connecting the sphere to earth provides a conducting path for some repelled electrons to leave the sphere. The sphere thereby acquires a positive net charge. Removing the earth connection while the rod remains nearby prevents the escaped electrons from returning. When the rod is finally withdrawn, the remaining positive charge redistributes over the sphere, but its positive net charge persists. In Experiment II, the earth connection remains present while the rod is withdrawn. With the external charged rod removed, the cause of the induced charge separation disappears. Electrons can flow back from earth until the conductor returns to the neutral equilibrium condition. Only after this redistribution is complete is the earth connection removed. The sphere therefore remains neutral in Experiment II. The decisive factor is the order of disconnecting the earth connection and withdrawing the charged rod. Electrostatic induction can produce a permanently charged conductor without physical contact with the inducing rod, but merely approaching and withdrawing the rod does not necessarily leave a net charge.

Explanation: \(\textbf{Initial total charge:}\) \[ Q_i=+5e-2e=+3e \] Since the system is isolated, every physically permissible final distribution must have algebraic total charge \(+3e\). In addition, each body's charge must be an integral multiple of \(e\), because electrons are transferred in whole numbers. \(\textbf{Examine Case I:}\) \[ Q_{\mathrm I}=4e-e=3e \] Both final charges are integral multiples of \(e\), and the total is conserved. The change corresponds to one electron moving from \(B\) to \(A\). \(\textbf{Examine Case II:}\) \[ Q_{\mathrm{II}}=\frac{5e}{2}+\frac{e}{2}=3e \] Although the total charge is conserved, neither listed final charge is an integral multiple of \(e\). This case violates microscopic charge quantisation. \(\textbf{Examine Case III:}\) \[ Q_{\mathrm{III}}=3e-e=2e \] The individual charges are quantised, but the final total differs from the initial total. This violates conservation of charge. \(\textbf{Examine Case IV:}\) \[ Q_{\mathrm{IV}}=e+2e=3e \] Both charges are quantised and their algebraic sum is conserved. This distribution corresponds to four electrons moving from \(B\) to \(A\). Therefore, Cases I and IV satisfy both necessary physical conditions. Conservation of total charge and quantisation of individual charges are separate tests; satisfying one does not automatically guarantee the other.

Explanation: \(\textbf{Contact rule:}\) Identical conducting spheres attain equal charges after being brought into contact and allowed to reach electrostatic equilibrium. For any two such spheres, \[ q_{\mathrm{each}}=\frac{q_1+q_2}{2} \] \(\textbf{Procedure I — first contact:}\) Sphere \(A\), carrying \(+6q\), touches neutral sphere \(B\). \[ q_A'=q_B'=\frac{6q+0}{2}=+3q \] Sphere \(C\) remains at \(-6q\) during this contact. \(\textbf{Procedure I — second contact:}\) Sphere \(A\), now carrying \(+3q\), touches sphere \(C\), carrying \(-6q\). \[ q_A''=q_C'=\frac{3q-6q}{2} \] \[ q_A''=-\frac{3q}{2} \] Thus, the final charge on \(A\) in Procedure I is negative. \(\textbf{Procedure II — first contact:}\) Initially, \(A\) carries \(+6q\) and \(C\) carries \(-6q\). \[ q_A'=q_C'=\frac{6q-6q}{2}=0 \] Both spheres become neutral after this contact. \(\textbf{Procedure II — second contact:}\) Sphere \(A\), now neutral, touches the initially neutral sphere \(B\). \[ q_A''=q_B'=\frac{0+0}{2}=0 \] \(\textbf{Conservation check:}\) The combined charge of all three spheres remains zero in both procedures, but their final individual charges differ because the contact sequence differs. \(\textbf{Final answer:}\) \[ \boxed{(q_{A,\mathrm I},q_{A,\mathrm{II}})=\left(-\frac{3q}{2},\,0\right)} \]

Explanation: \(\textbf{Initial interaction:}\) For two point charges separated by distance \(r\), Coulomb's law gives \[ F=k\frac{|q_1q_2|}{r^2} \] where \(k=1/(4\pi\varepsilon_0)\) in vacuum. Initially, \[ q_1=+4q,\qquad q_2=-q,\qquad r=2a \] Therefore, \[ F=k\frac{|(+4q)(-q)|}{(2a)^2} \] \[ F=k\frac{4q^2}{4a^2}=\frac{kq^2}{a^2} \] The initial force is attractive because the charges have opposite signs. \(\textbf{Modified interaction:}\) The charges become \(+4q\) and \(+q\), separated by \(a\). Their new force magnitude is \[ F'=k\frac{|(+4q)(+q)|}{a^2} \] \[ F'=\frac{4kq^2}{a^2} \] Comparing with the original magnitude, \[ \frac{F'}{F}=\frac{4kq^2/a^2}{kq^2/a^2}=4 \] The new force is repulsive because both charges are positive. \(\textbf{Newton's third-law check:}\) Each charge experiences the same magnitude \(4F\), with the force directions opposite to each other. Unequal charge magnitudes do not imply unequal mutual force magnitudes. \(\textbf{Final answer:}\) \[ \boxed{F'=4F\quad\text{(repulsive)}} \]

Explanation: \(\textbf{Initial Coulomb force:}\) In vacuum, \[ F=\frac{1}{4\pi\varepsilon_0}\frac{|q_1q_2|}{r^2} \] Let \(k=1/(4\pi\varepsilon_0)\). For the initial charges, \[ F=k\frac{|(+q)(-2q)|}{r^2} \] \[ F=\frac{2kq^2}{r^2} \] Because the signs are opposite, this force is attractive. \(\textbf{Effect of the dielectric:}\) In an ideal homogeneous dielectric medium of relative permittivity \(\varepsilon_r\), the Coulomb-force magnitude for fixed point charges is \[ F'=\frac{k}{\varepsilon_r}\frac{|q_1'q_2'|}{r'^2} \] Here, \[ q_1'=-2q,\quad q_2'=-3q,\quad r'=2r,\quad \varepsilon_r=3 \] \(\textbf{Substitution:}\) \[ F'=\frac{k}{3}\frac{|(-2q)(-3q)|}{(2r)^2} \] \[ F'=\frac{k}{3}\frac{6q^2}{4r^2} \] \[ F'=\frac{kq^2}{2r^2} \] \(\textbf{Force comparison:}\) \[ \frac{F'}{F}=\frac{kq^2/(2r^2)}{2kq^2/r^2} \] \[ \frac{F'}{F}=\frac14 \] Both modified charges are negative, so they repel each other. The new force is therefore repulsive despite being smaller in magnitude. \(\textbf{Final answer:}\) \[ \boxed{F'=\frac{F}{4}\quad\text{(repulsive)}} \]

Explanation: \(\textbf{Interpret the graph:}\) Coulomb's law in a homogeneous dielectric medium is \[ F=\frac{1}{4\pi\varepsilon_0\varepsilon_r}\frac{|q_1q_2|}{r^2} \] The horizontal graph variable is \[ x=\frac{1}{r^2} \] Therefore, \[ F=\left(\frac{k|q_1q_2|}{\varepsilon_r}\right)x \] This is a straight-line relation through the origin. \(\textbf{Meaning of slope:}\) The slope of the \(F\) versus \(1/r^2\) graph is \[ S=\frac{k|q_1q_2|}{\varepsilon_r} \] The charge product is unchanged between the two experiments. Hence the slope is inversely proportional to relative permittivity: \[ S\propto\frac{1}{\varepsilon_r} \] \(\textbf{Compare the two media:}\) \[ \frac{S_X}{S_Y}=\frac{\varepsilon_{r,Y}}{\varepsilon_{r,X}} \] The given slope ratio is \[ \frac{S_X}{S_Y}=\frac32 \] Since \(\varepsilon_{r,X}=2\), \[ \frac32=\frac{\varepsilon_{r,Y}}{2} \] \[ \varepsilon_{r,Y}=2\times\frac32=3 \] \(\textbf{Physical check:}\) Medium \(X\) has the steeper graph and therefore the smaller relative permittivity. The calculated value \(3\) for medium \(Y\) is greater than the given value \(2\) for medium \(X\), as expected. \(\textbf{Final answer:}\) \[ \boxed{\varepsilon_{r,Y}=3} \]

Explanation: \(\textbf{Initial force:}\) The initial charges are opposite in sign, so their interaction is attractive. Using Coulomb's law in vacuum, \[ F=k\frac{|q_1q_2|}{r^2} \] Substituting the initial charge magnitudes, \[ F=k\frac{(6\times10^{-6})(2\times10^{-6})}{r^2} \] \[ F=\frac{12k\times10^{-12}}{r^2} \] \(\textbf{Redistribution during contact:}\) The spheres are identical, so their final charges after contact are equal. Total charge is conserved: \[ Q=+6-2=+4\,\mu\text{C} \] Thus, \[ q_1'=q_2'=\frac{4}{2}=+2\,\mu\text{C} \] Both final charges are positive, so the new interaction will be repulsive. \(\textbf{Modified medium and separation:}\) The relative permittivity is \(2\), and the new separation is \(r/2\). Hence, \[ F'=\frac{k}{2}\frac{|q_1'q_2'|}{(r/2)^2} \] \[ F'=\frac{k}{2}\frac{(2\times10^{-6})(2\times10^{-6})}{r^2/4} \] \[ F'=\frac{k}{2}\frac{4\times10^{-12}\times4}{r^2} \] \[ F'=\frac{8k\times10^{-12}}{r^2} \] \(\textbf{Compare the two interactions:}\) \[ \frac{F'}{F}=\frac{8k\times10^{-12}/r^2}{12k\times10^{-12}/r^2} \] \[ \frac{F'}{F}=\frac23 \] The force magnitude decreases even though the separation is halved, because both charge redistribution and the dielectric medium affect the interaction. \(\textbf{Final answer:}\) \[ \boxed{F'=\frac{2F}{3}\quad\text{(repulsive)}} \]

Explanation: \(\textbf{Initial interaction:}\) Both original charges are positive, so the initial force is repulsive. By Coulomb's law, \[ F=k\frac{|(+Q)(+2Q)|}{r^2} \] \[ F=\frac{2kQ^2}{r^2} \] \(\textbf{Infer the new charge sign:}\) The second charge remains positive. The new measured force is attractive. Therefore, the replacement charge must be negative: \[ Q'\lt0 \] \(\textbf{Apply the changed separation:}\) The new separation is \(2r\), so \[ F'=k\frac{|Q'(2Q)|}{(2r)^2} \] \[ F'=\frac{2k|Q'|Q}{4r^2} \] \[ F'=\frac{k|Q'|Q}{2r^2} \] The observed force magnitude is \[ F'=\frac{F}{2} \] Using the original force expression, \[ F'=\frac12\left(\frac{2kQ^2}{r^2}\right) \] \[ F'=\frac{kQ^2}{r^2} \] \(\textbf{Determine the unknown magnitude:}\) Equating the two expressions, \[ \frac{k|Q'|Q}{2r^2}=\frac{kQ^2}{r^2} \] Cancelling common factors gives \[ |Q'|=2Q \] The attraction condition requires the negative sign. \(\textbf{Final answer:}\) \[ \boxed{Q'=-2Q} \] The result is physically consistent: doubling the charge magnitude compensates partly for the fourfold reduction in force produced by doubling the separation.

Explanation: \(\textbf{Force due to the left charge:}\) The central charge is negative, while the left charge is positive. They attract, so the force on the central charge is towards the left. The separation is \(a\). \[ F_L=k\frac{|(+2q)(-q)|}{a^2} \] \[ F_L=\frac{2kq^2}{a^2}=2F_0 \] Its vector direction is negative \(x\): \[ \vec F_L=-2F_0\hat i \] \(\textbf{Original force due to the right charge:}\) The initial right charge is \(+8q\), at distance \(2a\). Its attraction on the central negative charge is towards the right. \[ F_R=k\frac{|(+8q)(-q)|}{(2a)^2} \] \[ F_R=\frac{8kq^2}{4a^2}=2F_0 \] Thus, \[ \vec F_R=+2F_0\hat i \] The initial resultant is \[ \vec F_{\mathrm{initial}}=(-2F_0+2F_0)\hat i=0 \] \(\textbf{Effect of reversing the right charge:}\) The right charge becomes \(-8q\). Its magnitude and distance are unchanged, so the magnitude of its force remains \(2F_0\). However, the central charge and right charge are now both negative. They repel, pushing the central charge towards the left. \[ \vec F_R'=-2F_0\hat i \] \(\textbf{Apply superposition:}\) \[ \vec F_{\mathrm{net}}=\vec F_L+\vec F_R' \] \[ \vec F_{\mathrm{net}}=(-2F_0-2F_0)\hat i \] \[ \vec F_{\mathrm{net}}=-4F_0\hat i \] \(\textbf{Final answer:}\) \[ \boxed{\vec F_{\mathrm{net}}=-4F_0\hat i} \] The net force has magnitude \(4F_0\) and acts towards the negative \(x\)-direction. The important change is the reversal of one force direction, not its magnitude.

Explanation: \(\textbf{Set the force scale:}\) Let \[ F_0=\frac{kq^2}{a^2} \] The electrostatic forces must be calculated separately and added vectorially. \(\textbf{Initial force due to the charge at }(-a,0)\textbf{:}\) The source charge is \(-3q\), while the test charge is \(+q\). Their interaction is attractive, so the force acts along the negative \(x\)-direction. \[ F_x=k\frac{|(-3q)(+q)|}{a^2}=3F_0 \] Therefore, \[ \vec F_x=-3F_0\hat i \] \(\textbf{Force due to the charge at }(0,2a)\textbf{:}\) This source charge is \(+4q\), so it repels the positive test charge downwards. Its distance from the origin is \(2a\). \[ F_y=k\frac{|(+4q)(+q)|}{(2a)^2} \] \[ F_y=\frac{4kq^2}{4a^2}=F_0 \] Thus, \[ \vec F_y=-F_0\hat j \] \(\textbf{Initial resultant:}\) \[ \vec F_1=\vec F_x+\vec F_y \] \[ \vec F_1=F_0(-3\hat i-\hat j) \] Its magnitude is \[ |\vec F_1|=F_0\sqrt{(-3)^2+(-1)^2} \] \[ |\vec F_1|=\sqrt{10}F_0 \] \(\textbf{Resultant after charge reversal:}\) The source charge at \((-a,0)\) becomes \(+3q\). It now repels the positive test charge along the positive \(x\)-direction. Its force magnitude remains \(3F_0\). The force due to the upper charge remains unchanged. Therefore, \[ \vec F_2=F_0(3\hat i-\hat j) \] Its magnitude is also \[ |\vec F_2|=\sqrt{10}F_0 \] \(\textbf{Determine the angle:}\) Using the scalar-product relation, \[ \vec F_1\cdot\vec F_2=|\vec F_1||\vec F_2|\cos\theta \] Hence, \[ \cos\theta=\frac{\vec F_1\cdot\vec F_2}{|\vec F_1||\vec F_2|} \] Substituting the vector components, \[ \cos\theta=\frac{(-3F_0)(3F_0)+(-F_0)(-F_0)}{(\sqrt{10}F_0)(\sqrt{10}F_0)} \] \[ \cos\theta=\frac{-9F_0^2+F_0^2}{10F_0^2} \] \[ \cos\theta=-\frac{8}{10}=-\frac45 \] \(\textbf{Physical check:}\) The initial resultant points towards the lower-left quadrant, while the final resultant points towards the lower-right quadrant. Their horizontal components reverse, but their downward components remain unchanged. The angle between the resultants must therefore be obtuse, consistent with the negative cosine. \(\textbf{Final answer:}\) \[ \boxed{\cos\theta=-\frac45} \]

Explanation: Initially, the two touching conducting spheres form one electrically neutral conducting system. When the positively charged rod approaches sphere \(A\), the rod attracts mobile electrons within the connected conductor. Electrons move preferentially towards the nearer sphere \(A\), leaving a deficiency of electrons on the farther sphere \(B\). Thus, \(A\) develops negative net charge while \(B\) develops positive net charge. The total charge of the two-sphere system remains zero because there is no contact with the rod or earth. Separating the spheres while the rod is still nearby prevents electrons from flowing back between them. When the rod is subsequently withdrawn, the charges redistribute over the individual sphere surfaces, but their respective net charges remain unchanged. Therefore, \(A\) remains negative and \(B\) remains positive. During the final contact, the external charged rod is absent, and the total charge available to the connected pair is zero. Since the spheres are identical, they reach electrostatic equilibrium with equal final charges. Equal sharing of a zero total charge gives zero charge on each sphere. The important distinction is that induction initially separates charge without creating it, while subsequent contact allows the separated charges to neutralise.

Explanation: \(\textbf{Initial Coulomb interaction:}\) The two initial charges are \(+6q\) and \(-xq\). Their force magnitude is \[ F=k\frac{|(+6q)(-xq)|}{r^2} \] \[ F=\frac{6xkq^2}{r^2} \] \(\textbf{Charge redistribution during contact:}\) The spheres are identical, so they acquire equal charges after contact. Their conserved total charge is \[ Q=(6-x)q \] Therefore, each final charge is \[ q_f=\frac{(6-x)q}{2} \] The positive-probe observation establishes that both final charges are positive. Hence, \[ 6-x\gt0 \] \[ x\lt6 \] \(\textbf{Force after separation:}\) The separation returns to \(r\), so \[ F'=k\frac{q_f^2}{r^2} \] Substituting the redistributed charge, \[ F'=k\frac{(6-x)^2q^2}{4r^2} \] The measured ratio is \[ \frac{F'}{F}=\frac13 \] Thus, \[ \frac{(6-x)^2}{24x}=\frac13 \] \[ (6-x)^2=8x \] Expanding, \[ 36-12x+x^2=8x \] \[ x^2-20x+36=0 \] Factorising, \[ (x-2)(x-18)=0 \] Therefore, \[ x=2\quad\text{or}\quad x=18 \] \(\textbf{Select the physically admissible root:}\) The mathematical solution \(x=18\) would produce \[ q_f=\frac{(6-18)q}{2}=-6q \] Both spheres would then be negatively charged, contradicting the positive-probe observation. The condition \(x\lt6\) selects \(x=2\). For this value, each sphere finishes with charge \(+2q\). The final force ratio is \[ \frac{F'}{F}=\frac{(2q)(2q)}{(6q)(2q)}=\frac13 \] This verifies the measured force ratio and the final positive signs. \(\textbf{Final answer:}\) \[ \boxed{|q_{\mathrm{initial,negative}}|=2q} \]

Explanation: \(\textbf{Mutual electrostatic force:}\) The charges have opposite signs, so their interaction is attractive. At the instant of release, their separation is \(r\). The magnitude of the force on either particle is \[ F=k\frac{|(+q)(-4q)|}{r^2} \] \[ F=\frac{4kq^2}{r^2} \] Newton's third law requires the mutual forces to have equal magnitudes and opposite directions. The unequal charges do not produce unequal mutual force magnitudes. \(\textbf{Acceleration of particle }A\textbf{:}\) Using Newton's second law, \[ a_A=\frac{F}{m_A} \] \[ a_A=\frac{4kq^2/r^2}{2m} \] \[ a_A=\frac{2kq^2}{mr^2} \] \(\textbf{Acceleration of particle }B\textbf{:}\) Similarly, \[ a_B=\frac{F}{m_B} \] \[ a_B=\frac{4kq^2/r^2}{5m} \] \[ a_B=\frac{4kq^2}{5mr^2} \] \(\textbf{Compare the accelerations:}\) \[ \frac{a_A}{a_B}=\frac{F/(2m)}{F/(5m)} \] \[ \frac{a_A}{a_B}=\frac52 \] Both acceleration vectors point towards the other particle because the interaction is attractive. Their acceleration directions are opposite in a common coordinate system. \(\textbf{Physical check:}\) Particle \(A\) has the smaller mass, so it must have the larger acceleration under the same force magnitude. This agrees with the calculated ratio. \(\textbf{Final answer:}\) \[ \boxed{a_A:a_B=5:2\quad\text{(mutually attractive)}} \]

Explanation: \(\textbf{Identify the governing distance law:}\) For fixed point charges in vacuum, \[ F=k\frac{|q_1q_2|}{r^2} \] Thus, \[ Fr^2=k|q_1q_2|=\text{constant} \] The measurement at separation \(a\) gives \[ F(a)=36\,\text{N} \] Therefore, the expected force at an arbitrary separation \(r\) is \[ F(r)=36\left(\frac{a}{r}\right)^2\text{N} \] \(\textbf{Check the observations:}\) At separation \(2a\), \[ F(2a)=36\left(\frac12\right)^2=9\,\text{N} \] This agrees with the recorded value. At separation \(3a\), \[ F(3a)=36\left(\frac13\right)^2 \] \[ F(3a)=4\,\text{N} \] The recorded value is \(5\,\text{N}\), so this measurement is inconsistent with the inverse-square law. At separation \(4a\), \[ F(4a)=36\left(\frac14\right)^2 \] \[ F(4a)=\frac{36}{16}=2.25\,\text{N} \] This agrees with the table. \(\textbf{Apply the dielectric correction:}\) The incorrect measurement corresponds to separation \(3a\). The correct vacuum force at this separation is \(4\,\text{N}\). For a homogeneous dielectric of relative permittivity \(\varepsilon_r=2\), \[ F_{\mathrm{medium}}=\frac{F_{\mathrm{vacuum}}}{\varepsilon_r} \] \[ F_{\mathrm{medium}}=\frac{4}{2}=2\,\text{N} \] \(\textbf{Physical check:}\) The dielectric reduces the force magnitude for the same charge configuration. The expected force must therefore be smaller than the corrected vacuum value of \(4\,\text{N}\). \(\textbf{Final answer:}\) \[ \boxed{F_{\mathrm{medium}}=2.0\,\text{N}} \]