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In Fig. 25-3, the potential difference between the metal plates in air is $40 \mathrm{~V}$. (a) Which plate is at the higher potential? (b) How much work must be done to carry a $+3.0$ C charge from $B$ to $A$ ? From $A$ to $B$ ? (c) How do we know that the electric field is in the direction indicated? $(d)$ If the plate separation is $5.0 \mathrm{~mm}$, what is the magnitude of $\overrightarrow{\mathbf{E}}$ ? (a) A positive test charge between the plates is repelled by $A$ and attracted by $B$. Left to itself, the positive test charge will move from $A$ to $B$, and so $A$ is at the higher potential. (b) The magnitude of the work done in carrying a charge $q$ through a potential difference $V$ is $q V$. Thus the magnitude of the work done in the present situation is $$ W=(3.0 \mathrm{C})(40 \mathrm{~V})=0.12 \mathrm{~kJ} $$ Because a positive charge between the plates is repelled by $A$, positive work $(+120 \mathrm{~J})$ must be done to drag the $+3.0 \mathrm{C}$ charge from $B$ to $A$. To restrain the charge as it moves from $A$ to $B$, negative work $(-120 \mathrm{~J})$ is done. (c) A positive test-charge between the plates experiences a force directed from $A$ to $B$ and this is, by definition, the direction of the field. (d) For closely spaced parallel plates, $V=E d$. Therefore, $$ E=\frac{V}{d}=\frac{40 \mathrm{~V}}{0.0050 \mathrm{~m}}=8.0 \mathrm{kV} / \mathrm{m} $$ Notice that the SI units for electric field, $\mathrm{V} / \mathrm{m}$ and $\mathrm{N} / \mathrm{C}$, are identical.

Hi There are three charges yep. At the vortices of Anita electoral triangle like this. The three charged particles are having the values upper one. This is plus 4.0 micro column. The lower one the left lower less 2.0 Micropal. Um And the right lower that is Plus 3.0 Micro column. Let it be A. The lower left that is represented by B. And the lower right support. This is C. Each side of this triangle is 20 centimetre. Cool. This upper charge Will be experiencing two courses of repulsion. This upper one upper charge Plus 4.0 micro colon. It will be repelled by the charge capped at B means plus 2.0 micro colon. And it will be repelled by the charge kept at sea means plus 3.0 Mike local. I'm also these two forces maybe named as F. A. B. Force experienced by the charge put at A. Due to that. Could that be? And F. A. C. Most experienced by the charge put at a. Due to that put at sea. And the angle between them will be 0°. Which will be helpful to find their vector addition. Now using columns law. F A. C will be given by a. Into you A. Into Qc. Charge put at A. And C. Respectively divided by the square of distance between them which is nothing but the side of his square In m. This will be 0.2 meter. The whole square. Now putting the values all other known values for K. This is nine in 2 10.9 into for cure a disease for micro problem or four into 10 for -6 column into QC which is three into 10 days par -6. School. I'm divided by 0.04. So finally this force comes out to be so 0.7 Newton. Similarly, F. A into Q. B. Invited by saying the side of this Equal Little Triangles 0.2 m. Or it will give it will be given by nine in 2. And as to the part nine, You took you a again four into 10 days par -6. You'll be To into 10 par -6 Divided by Square of 0.2, which will come out to be 0.0 both. And finally this force. F A B comes out to be 1.8 newton and angle between them That is 60°. Using the concept of vertically opposite angles. Well, finally, the net force acting at this charge put at the vortex A. Using triangle law of tradition, it will be given by square root of squared off FCC classes quite off. F A B plus two times of F A. C into F A B into cosine of angle between them which is 60° and cost 60° comes out to be half. Now putting all these values here Or FAC. This is 2.7. Newton Have a baby that is one pointed newton Last two times off 2.7- 1.8, Cost 60° half cancelling this half here. Finally, we get This is 7.29 less, 3.24 plus 4.86, which comes out to be equal to 15 point 39 Newton. Or finally, magnitude of this net force Comes out to be 3.9 two newton, which is the answer for the given problem here. Thank you.

Everybody. So look in it to solve. Serious for the cap, since for a so a we have C equals C wine A plus C two and C one is gonna equal to K. Absalon. Ah, a divided by D The area has been equal LX okay, And then, um, area to is gonna equal to Oh, minus x times x. Okay, so for this we have k Absolutely not times. Um oh, minus x x divided by D plus K. Absalon l X Divided by D a Khiry absolute not l divided by d times l plus, um K minus one times X. Okay. Yeah, And let's just make sure here at what we got k From there we go. Okay, eyes. And now for Herbie we're finding potential. So yet u equals c v squared so we could do one to you One time's to absolute not oh v squared divided by D Okay, and we do OK sti um times l plus K minus one x um And for this one, we have a equals. Um, you take the derivative of the eggs. So because of that, we have d u equals absolutely not l b squared two d k minus one. Okay, d x and would set the limits as x two x and x two x plus d x of a and, um, we get d u equals u X plus the x minus You x Okay, Yeah, And the word is gonna bring it down and the absolute not I'll be squared. Two D l plus K minus one x plus d x minus. Absalon not L V squared two D l waas K minus one x. Okay. And we're just doing a short cut version and you can do this. I'm sure your capitalist professor has found of this. We're just doing a very short cut. Okay, An absolute not l squared V squared two D plus Absalon, not L V squared two D k minus one, eggs plus absolute not l V squared two D K minus one d X. Then we have minus well, to close their. Then we have, um, minus absalon, not l b squared two D K minus one x. Okay. And this is gonna be equal to absolute, not l B squared, divided by two D. K minus one D X five and we're still not quite dying. We have store more parts to figure out for this. So we have Q equals See one V. And so it's gonna be K epsilon, not l x V divided by D for the first slab. And then, um, we have k to the epsilon, not o minus x x times V um de and this is gonna be up salon, not our minus x x b d. Ok, And now for, um, the Q total you got que total. Okay, you got CV, which is gonna equal tau epsilon not l de and we'll put locates d here. So D l plus, um K minus one x times leave. Okay. And now vying at the potential so can be cute Teoh. See, Then we at D you over d c equals negative. Um, que squared. They're gonna give to, um and Q squared two c squared. Do you d x over d X equals negative. Hugh squared to see squared D u d x times d x If I am so, your d u is going to be negative, you squared times two c squared and we're gonna put what we found. A boat. It's absolutely not. I'll be squared. Two d. K minus one. The X. Okay. And, um yeah, And then, um that is your answer. And now, word for tea. Um, when you find the force so got D u equals, give the forced the X Eagles. Absolutely not. L the squared to de yeah, sooty K minus wine the X And then we have after the axe equals Don't take the derivative of this, So we're actually going to get, um, a B is gonna equal to negative. Absolute, not L b squared two D A minus one. Okay. And for up. See, we're basically gonna do the same thing He squared. Donna l two D. Um, K minus one, the x and, um, you're and then, um, for this, your app is gonna upsy is unequal to be squared. Absolute, not l Serie de K minus one. Okay. And it e is your force, and your force is going to be positive. Is the Epson on not l V squared two d K minus one. Okay, guys. Thank you.

Everyone. As soon as the figure, there will be two cap distress even in C two and David Wynn Palace. So she's cool to see even Placido. She even is. Absolutely not. Yeah. Okay, here. I have to make a correction. Mhm. This is given X and this will be l minus six. Yeah, okay. Had into X over the It is absolutely not. And into all minus x mhm abandon. So this is a totally tapestries of body. In part, we we have to find them changing internal energy. There you is. Card to have d c into the square here D c is to find as d. Absolutely not. End K minus one. Yeah, into X. Okay. Us and yeah, apart. Indeed. Mhm. Mm. Mhm. So you will get a disease called to Absolutely not. And here, my husband d x upon the so energy store, baby. Absolutely not. We square head. Okay. Minus one D x. Mm. Upon pull it. Yeah. If okay, due to be cause kept constant, that is. Caps traceable. Increase. Okay, So energy very decrease. Yeah. Has energies to find us Jewish, Whereby to see Okay, Mhm. And the force is given by minus. Do you? Upon the it's that is Do you expect her to minus FDX? So force is approaching the Yeah. Motion? Yeah. Equality. It happens because system I would like to Yeah, okay. Uh huh. Minimise its energy. Okay. Yeah. Oh, yeah. There. Thanks.


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