This page breaks down an important question from the Class 12 Physics NCERT Exemplar book, offering a full, step-by-step solution. It hones in on Chapter 2, Electrostatic Potential and Capacitance. The real value here? These exemplar problems push you to actually grasp the concept, not just skim the surface—they dig deep.
Is question mein hum electric potential difference, point charges, aur work done ke concepts ko ek saath use karenge. Dekhte hain, pehle basic idea clear kar lein—phir is numerical ko step-by-step solve karte hain, bilkul aaram se.
Here we have provided NCERT Exemplar Questionfor Class 12 Physics in hindi Language, Just select the chapters below to get Exemplar Solution of the same:
Question: Two point charges, +4 μC and -2 μC, are sitting 10 cm apart from each other. On the line that joins them, where exactly does the electric potential drop to zero? (And yeah, we’re taking potential at infinity as zero, as usual.) So you’ve got these two charges, one positive and one negative. They’re not too far off—just 10 cm. The trick here is to find that sweet spot on the straight line between (or beyond) them where their potentials cancel out. Remember, potential isn’t a vector. It’s just a straight sum—no direction drama. One charge pulls it up, the other pulls it down, and somewhere along that line they balance each other to zero. Here’s the thing though—it might not just be between them. Could be on either side, too. You’ve got to check both regions: outside the pair on the positive side, outside on the negative side, and yeah, in the middle as well. But don’t overthink it. Start by writing out the potential from each charge at a general point, set their sum to zero, and solve for the distance. One of those spots will pop out cleanly. The other might be trickier, depending on the magnitudes. Just keep the signs straight—that +4 and -2 matter a lot. And since infinity’s already zero, you’re basically hunting for where the two contributions cancel dead even. That’s your answer.
This question? It’s straight off the superposition principle for electric potential. Plain and simple. At any given point, you just add up the potential from each individual charge—algebraically, like you’re summing numbers that can be positive or negative. That’s it — no vector gymnastics here, just direct addition. The formula you’ll lean on is V = kq/r, with k being that constant, 1/(4πε₀). So whenever you see multiple charges hanging around, don’t overthink it. Add their potentials, keep the signs straight, and you’re good.
Okay, so first things first—Step 1. You gotta really get the situation before you do anything else. I mean, truly understand it. Don't just skim the surface and think you've got it. That's a trap. Look at what's right in front of you, sure, but also dig a little deeper. What's the actual problem here? Who's involved? What's at stake? Get a clear, honest picture of where things stand right now. It's like checking the map before you start driving—you wouldn't just hit the gas and hope for the best. So take that beat, slow down, and really wrap your head around the scenario you're dealing with.
Okay, so we’ve got a +4 μC charge sitting right at the origin, point O. Then, over on the x-axis, 10 cm away, there’s a -2 μC charge. What we need is to pin down that exact spot where the total electric potential just cancels out to zero. Let’s call that point P. Say it’s some distance, we’ll call it x, measured in meters straight from the +4 μC charge, and it’s got to be right on that same line connecting them. That’s our starting assumption.
Step 2’s all about setting up the potential Ka equation. Honestly, it’s not as scary as it sounds. You’re basically just writing down the acid dissociation expression—Ka equals the concentration of your products over the reactants. And don’t forget to square things if you’ve got a diprotic acid or something with stoichiometry that calls for it. Just lay it out clean, plug in what you know later. You’re golden. It’s the foundation, so take your time here—it makes the next steps way smoother.
Start with the charges themselves. The +4 μC pushes its potential out into the space around it, and the -2 μC does the same, just in the opposite direction. Now, at point P, you add those two contributions up. That's the whole trick. V_total equals the potential from the +4 μC plus the potential from the -2 μC. And when you actually do that sum, it lands exactly on zero. Both fields cancel each other out right there. Clean, right? That's your answer.
Iska matlab yeh hai ke: k times 4 x 10⁻⁶, sab x ke upar, plus k times -2 x 10⁻⁶, sab (0.10 - x) ke upar, aur yeh sab zero ke barabar hai. Bas yahi hai.
Well, now we get to the real work—Step 3 is where you actually solve the equation. No more setting things up or rearranging terms; this is the moment you roll up your sleeves and crunch the numbers. You’ll want to isolate that variable, do the math step by step, and keep your work tidy so you can spot any slip-ups. Honestly, this part can get a little messy sometimes. That’s totally normal. Just take it one move at a time, and check each line as you go. The goal’s simple: get the variable alone on one side, and you’re basically there.
Okay, so the 'k' just cancels out. Poof, gone. You're left with (4/x) - (2/(0.10 - x)) = 0. That's the equation you're working with now.
Aur yahi cheez humein milti hai: 4/x = 2/(0.10 - x). Bas, itna hi.
Cross multiply karte hain — straightforward hai. 4 times (0.10 minus x) equals 2 times x. Bas yahi hai.
0.40 minus 4x equals 2x — simple enough, right? Just gotta get those x terms on the same side.
0.40 equals 6x — that’s it. No tricks, no extra steps—just a straight-up equation staring you in the face. And honestly, you could solve it in your head if you wanted to. Divide both sides by 6, and you’re done. Boom. 0.40 divided by 6 gives you that tiny, awkward decimal—about 0.0667, if you’re keeping score. But the real point here? The numbers don’t lie. You’ve got a product, a multiplier, and a fixed result. Six times something, and that something is small. Real small. So yeah, 0.40 is six times that value, plain and simple.
Okay, here’s that rewrite: So, you take 0.40 and divide it by 6, which gives you 0.0667. That’s meters, by the way. Convert that over to centimeters, and you land on 6.67 cm. Simple as that.
Step 4 is where things actually come together. You’ve done the work, run the numbers, and now you’re staring at the result—so what’s the story? This part’s all about reading between the lines and figuring out what that outcome really means for you. Don’t just glance at it and move on. Sit with it for a second. Does it match what you expected, or did it throw you a curveball? Interpret it in plain terms, not jargon. That’s the whole game here—turning raw output into something you can actually use, and honestly, that’s where the value lives. It’s not just a final number or a done deal. It’s a clue. And once you decode it, you’re not stuck at the finish line—you’re ready to make your next move.
Dekho, baat samajhni hai toh seedhi si hai. Zero potential ka point +4 μC wale charge se 6.67 cm door hoga, bilkul dono charges ke beech mein. Aur yeh galat nahi hai, kyunki jo charge positive hai woh kaafi bada hai. Isliye zero potential wali jagah naturally negative charge ki taraf shift ho jaati hai. Samajh gaye na?
CBSE ke pattern mein, aise numericals usually 3-4 marks ke short answer type questions mein poochhe jaate hain. Agar concept clear ho, toh phir derivations bhi simple lagne lagte hain—aur MCQs toh chhote bachche ka khel lagte hain. Electrostatic potential ka yeh basic application aapko capacitance ke numericals mein bhi kaam aayega, aur wahan aapko feel hoga ki base kitna strong hai.
Keep this in mind, okay? Always cross-check your answer—does it actually make sense for the physical situation you're dealing with? Because here's the thing, a negative distance just isn't a thing. That won't happen. And your point charges — their positions have to fall somewhere in between. That's the only place they can be. So if you get something that breaks those rules, you've probably messed up somewhere along the line.
