Wow, sir you are the only one who teaches wih intuition. And intuition makes things easy, you gave me the idea of what electric field is and gave the intuition. For me, you teach even better than physicswallah
Hey can we direct say for work done its force times displacement for every distance we move forward isn't the force gonna be less and we have to do integral
How can Work's unit be (J/C) And then you put it equals to (V) whhich is unit of Potential but we have to found Pitential Energy ???? I love Physics please help by answering these Questions By thw way Sir your voice is soothing
Electric potential is the amount of work done PER UNIT CHARGE to bring it from infinity to a particular point against the direction of electric field. Like how we take distance per unit time is speed, or mass per unit volume is density, similarly. And they have put 1J/1C= 1V 1 volt. A new name.
He should also divide W (not value,that letter ,,.On left side as u can see))by q That is E=100N/C ~ [Work and potential energy have same equation (kq1q2)/r So he chose *WORK] Work /qr =E ~[But 1st he only divide q but not written on left hand side] Work /q=V since q=1 Work/1=V Work =V Work = potential Then divide the answer V is Divided by r ,so we get V/r= E Since r is also 1 V/1=E V=E Potential=Electric field strength Since work=potential=electric field strength V =100Er E=100V/r Hope it will help u 🥰🇮🇳
Sir you said right side is positive when you are moving positive charge toward positive then you have to done work and then as we are moving toward high potential then we should gain p.E as when we move up we gain P.E????
@@tiwariji8176 well since the charge is moving in the direction of where force is being applied by the field lines it starts to accelerate and looses potential energy which gets converted to kinetic energy and by work energy theorm the work done by it to move is the enrgy lost HOPE IT HELPS!
@@shriyasachdeva1024is this relation of E = V/meter , only applicable for Uniform field lines ? Or it is valid for any electric field i.e. for variable Electric field ?
@@Nancy_Shorts13 if electric field is non uniform dv/dr is used if you solve numericals you will understand if electric field is varible we cant just write its mag until we talk about a point
Negative charge goes from lower potential to higher potential. So the change in potential will be positive. In that case, change in r is negative, meaning the separation will decrease, if the reference charge is positive. However, if the reference charge is also negative, then from the formula E = -grad(V), we get a negative magnitude of E. But how can the 'magnitude' of a vector be negative?
It's crazy that this is free, but I have to pay for a college professor to just send me to Pearson's Mastering Physics, where I read and try to understand everything by myself. No lectures, or powerPoints, and bare minimum communication. American education is a scam
*_Summary Of The Video :-_* *Relation Between Electric Field And Potential* * The Relation define the Electric Field in term of potential. * Unit - Volt / meter. * It Is The Measure Of How Much Potential Drops Per Meter As We Go Forward In The Direction Of Electric Field. * Negative Sign Represents How Much Potential Drops. * Whenever something is changing with respect to distance it is known as *Gradient* . * In above case potential is changing with respect to distance So it is known as *Potential Gradient* *Definition -* "Electric Field Is The Negative Potential Gradient"
but, you are just equation those two quantities, potential ( in its essence) defined as the " concentratio of positive charges" and the electric force is something thats different than the concentration of postive charges, why do you equate them without any logical implication , i mean , i can assume that E = dV/dx but , i dont see, the logical implication , or is it some other law of physics that relates the electric force with the gradient of concentration of electric charge?, greetings
