The electric potential at a given position...
{"title":"The electric potential at a given position...","description":"Correct!","type":"correct","color":"#99CCFF","code":"[{\"shape\":\"poly\",\"coords\":\"82,133\"},{\"shape\":\"rect\",\"coords\":\"10,16,12,16\"},{\"shape\":\"poly\",\"coords\":\"144,22\"},{\"shape\":\"rect\",\"coords\":\"3,46,48,98\"}]"} {"title":"…equals the electric potential energy for a charge q sub 0 at that position…","description":"Wrong","type":"incorrect","color":"#ffcc00","code":"[{\"shape\":\"rect\",\"coords\":\"111,7,154,52\"}]"} {"title":"…divided by the charge q sub 0.","description":"Incorrect","type":"incorrect","color":"#333300","code":"[{\"shape\":\"rect\",\"coords\":\"227,24,228,24\"},{\"shape\":\"rect\",\"coords\":\"163,104,201,149\"}]"}Our discussion in Section 17-2 shows that if a point charge q changes position, the potential energy change Uelectric depends on both the magnitude and the sign (positive or negative) of q (see Figure 17-3). We can simplify things by considering the potential energy per charge—that is, the electric potential energy for a charge at a given position divided by the value of that charge. We call this quantity the electric potential and denote it by the symbol V:
We call the charge q0 the test charge: Its charge has such a small magnitude that it doesn’t affect the other charges that create the electric field in which q0 moves. Because we divide out the value of q0, the value of the potential V at a given position does not depend on the value of the point charge q0 that we place there. Instead, V is determined by the other charges that produce the electric field at the position where we place the test charge.