4 µC के किसी आवेश को दो भागों में विभाजित किया जाना है। यदि विभाजित आवेशों के बीच दूरी नियत रखी जाए तो विभाजित आवेशों का परिमाण क्या होगा, जिसके लिए उनके बीच बल अधिकतम होगा -

Physics
4 µC के किसी आवेश को दो भागों में विभाजित किया जाना है। यदि विभाजित आवेशों के बीच दूरी नियत रखी जाए तो विभाजित आवेशों का परिमाण क्या होगा, जिसके लिए उनके बीच बल अधिकतम होगा -

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Answer

(b) 2 μC और 2 μC

Step 1: The electrostatic force between two point charges q1q_1 and q2q_2 separated by a fixed distance rr is given by

F=k∣q1q2∣r2F = \frac{k |q_1 q_2|}{r^2}

where kk is Coulomb's constant. Here, q1+q2=4 μCq_1 + q_2 = 4\,\mu\mathrm{C} and rr is fixed, so F∝∣q1q2∣F \propto |q_1 q_2|.

Step 2: Let q1=q μCq_1 = q\,\mu\mathrm{C} and q2=(4−q) μCq_2 = (4 - q)\,\mu\mathrm{C}. Then, the product is

P=∣q(4−q)∣ \muC2=∣4q−q2∣ \muC2.P = |q(4 - q)|\,\muC^2 = |4q - q^2|\,\muC^2.

Since 0≤q≤40 \leq q \leq 4, P=4q−q2P = 4q - q^2.

Step 3: To maximize PP, take the derivative:

dPdq=4−2q.\frac{dP}{dq} = 4 - 2q.

Set dPdq=0\frac{dP}{dq} = 0:

4−2q=0  ⟹  q=2.4 - 2q = 0 \implies q = 2.

Step 4: Second derivative test:

d2Pdq2=−2<0,\frac{d^2P}{dq^2} = -2 < 0,

so maximum at q=2 μCq = 2\,\mu\mathrm{C}.

Step 5: Verify with options (products in \muC2\muC^2):
(a) 1×3=31 \times 3 = 3
(b) 2×2=42 \times 2 = 4 (maximum)
(c) 0×4=00 \times 4 = 0
(d) 1.5×2.5=3.751.5 \times 2.5 = 3.75

(b) 2 μC और 2 μC

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The electrostatic force between two point charges q_1 and q_2 separated by a fixed distance r is given by F = (k |q_1 q_2|)/(r^2) where k is Coulomb's constant.

4 µC के किसी आवेश को दो भागों में विभाजित किया जाना है। यदि विभाजित आवेशों के बीच दूरी नियत रखी जाए तो विभाजित आवेशों का परिमाण क्या होगा, जिसके लिए उनके बीच बल अधिकतम होगा -
Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

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Step 1: The electrostatic force between two point charges q_1 and q_2 separated by a fixed distance r is given by F = (k |q_1 q_2|)/(r^2) where k is Coulomb's constant. Here, q_1 + q_2 = 4\, and r is fixed, so F |q_1 q_2|. Step 2: Let q_1 = q\, and q_2 = (4 - q)\,. Then, the product is P = |q(4 - q)|\,^2 = |4q - q^2|\,^2. Since 0 ≤ q ≤ 4, P = 4q - q^2. Step 3: To maximize P, take the derivative: (dP)/(dq) = 4 - 2q. Set (dP)/(dq) = 0: 4 - 2q = 0 q = 2. Step 4: Second derivative test: (d^2P)/(dq^2) = -2 < 0, so maximum at q = 2\,. Step 5: Verify with options (products in ^2): (a) 1 × 3 = 3 (b) 2 × 2 = 4 (maximum) (c) 0 × 4 = 0 (d) 1.5 × 2.5 = 3.75 (b) 2 μC और 2 μC