If the charge on the capacitor is increased by 2 coulomb, the energy stored in it increases by 21%. The original charge on the capacitor is:A. 10CB. 20CC. 30CD. 40C

We can solve this problem using the formula for the energy stored in a capacitor. The energy stored in a capacitor is given by:
U=Q22CU = \frac{Q^2}{2C}
Where:
• UU is the energy stored in the capacitor,
• QQ is the charge on the capacitor,
• CC is the capacitance of the capacitor.
Step 1: Let the original charge on the capacitor be Q1Q_1.
The energy stored in the capacitor with charge Q1Q_1 is:
U1=Q122CU_1 = \frac{Q_1^2}{2C}
Step 2: After increasing the charge by 2 coulombs, the new charge becomes Q2=Q1+2Q_2 = Q_1 + 2.
The new energy stored in the capacitor is:
U2=Q222C=(Q1+2)22CU_2 = \frac{Q_2^2}{2C} = \frac{(Q_1 + 2)^2}{2C}
Step 3: According to the problem, the energy increases by 21%. This means:
U2=1.21U1U_2 = 1.21 U_1
Substitute the expressions for U2U_2 and U1U_1:
(Q1+2)22C=1.21⋅Q122C\frac{(Q_1 + 2)^2}{2C} = 1.21 \cdot \frac{Q_1^2}{2C}
Step 4: Simplifying the equation:
(Q1+2)2=1.21Q12(Q_1 + 2)^2 = 1.21 Q_1^2
Expanding both sides:
Q12+4Q1+4=1.21Q12Q_1^2 + 4Q_1 + 4 = 1.21 Q_1^2
Rearranging the terms:
Q12−1.21Q12+4Q1+4=0Q_1^2 - 1.21 Q_1^2 + 4Q_1 + 4 = 0 −0.21Q12+4Q1+4=0-0.21 Q_1^2 + 4Q_1 + 4 = 0
Step 5: Solve this quadratic equation:
We can multiply through by 100100 to avoid decimals:
−21Q12+400Q1+400=0-21 Q_1^2 + 400 Q_1 + 400 = 0
Now solve the quadratic equation using the quadratic formula:
Q1=−b±b2−4ac2aQ_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=−21a = -21, b=400b = 400, and c=400c = 400.
Q1=−400±4002−4⋅(−21)⋅4002⋅(−21)Q_1 = \frac{-400 \pm \sqrt{400^2 - 4 \cdot (-21) \cdot 400}}{2 \cdot (-21)} Q1=−400±160000+33600−42Q_1 = \frac{-400 \pm \sqrt{160000 + 33600}}{-42} Q1=−400±193600−42Q_1 = \frac{-400 \pm \sqrt{193600}}{-42} Q1=−400±440−42Q_1 = \frac{-400 \pm 440}{-42}
We get two solutions:
Q1=−400+440−42=40−42=−2021(not a valid solution)Q_1 = \frac{-400 + 440}{-42} = \frac{40}{-42} = -\frac{20}{21} \quad (\text{not a valid solution})
and
Q1=−400−440−42=−840−42=20 CQ_1 = \frac{-400 - 440}{-42} = \frac{-840}{-42} = 20 \, \text{C}
Thus, the original charge on the capacitor is 20 C.
Final Answer:
The original charge on the capacitor is 20 C.
So, the correct option is B. 20 C.