Two charged particles are fixed to an x axis, 91 = -60uC is at x = +30cm and
+60uC is at x2 = -30cm. Find the
electric field at a point on the y axis at y
40cm?
Select one:
O +1,3 x 10 NFC
0 -2.6 x 106 î NC
0 -1.3 x 106 î N/C
O +2.6 x 106 î N/C
O -1.3 x 10° N/C

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

26 is 33.33% of 78

step-by-step explanation:

26 is what percent of 78?

26 is p% of 78

equation: y = p% * x

solving our equation for p

p% = y/x

p% = 26/78

p = 0.3333

convert decimal to percent:

p% = 0.3333 * 100 = 33.33%

4

step-by-step explanation:

im pretty sure it's 4 beacsue 8/2 equals 4

do have an equation or graph or

step-by-step explanation: