A positive charge sits at the origin. Where is the electric potential (V) minimu

Question

A positive charge sits at the origin. Where is the electric potential (V) minimum? At the origin At infinity Somewhere in between Can't determine The diagram shows four pairs of large parallel conducting plates. The value of the electric potential is given for each plate. Rank the pairs according to the magnitude of the electric field between the plates, least to greatest. 1, 2, 3, 4 4, 3, 2, 1 2, 3, 1, 4 2, 4, 1, 3 3, 2, 4, 1 Consider the three resistors and the battery in the circuit shown. Which resistors, if any, are connected in serial? R_2 and R_1 R_2 and R_3 R_1 and R_3 R_1 and R_2 and R_3 This cannot be determined Two capacitors A and B are connected in parallel for a long time and C_A = 2C_B. How does the charge on capacitor A compare to that on B? q_A = q_B/4 q_A = q_B/2 q_A = q_B q_A = 2q_B q_A = 4q_B If the charge on a parallel-plate capacitor is doubled: the capacitance is halved the capacitance is doubled the electric field is halved the electric field is doubled the surface charge density is not changed on either plate If the potential difference across a resistor is halved:

Explanation / Answer

(1) Electric potential is minimum at infinity .option (B)

Explantion : we have electric potential V=KQ/r

V=0 when r=infinity .

(2)

We have net potential at

1st plate =70-20=50V

2nd plate =(70+20)=90V

3rd plate =(90-10)=80V

4 rh plate =(90+130)=220

Now higher the potential higher is the electric field.

so order is from lowest highest

1<3<2<4

option none of these.

(4)

Charge on capacitor 1 is qA=(CB/(CA+CB))Q

where Q is total charge .

charge on capacitor 2 qB=(CA/(CA+CB))Q

Now qA/qB=CB/CA

=CB/2CB   (CA=2CB)

=1/2

qA=qB/2  ANS

Option (B)

(5)

we have Q=CV (capacitance will remain same only voltage will change)

here V=Q/C

Now when charged is doubled 2Q=CV'

Here V'=2Q/C

as the potential is doubled so electric field is also doubled .

option D ANS

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