Question

Three pairs of large conducting plates are charged to different potentials as shown in figure. The separation between each pair of plates is the same. The pair of plates according to the magnitude of the electric field between them, from highest to the least is

Answer 1

In order to solve this question, we will first find the net potential difference between these two plates in each of given case and then using general relation between electric field and potential difference we will figure out the relation between electric field strength in each of three cases.

Complete Step By Step Answer:
According to the given diagram we can see that,
for pair of plates in case $(1)$ the net potential difference is just the difference between higher and lower potential of plates so potential difference between these plates is ${V_1} = 80 - ( - 30)$ so, ${V_1} = 110V$
Similarly, for case $(2)$ the potential difference between these plates can be calculated as ${V_2} = 70 - 20$ so, ${V_2} = 50V$
Similarly for case $(3)$ the potential difference between these plates can be calculated as ${V_3} = 90 - ( - 10)$ so, ${V_3} = 100V$
Now, as we know that potential difference V, electric field E and separation between the potential plates d is related as $V = Ed$ so, as distance between the plates is same in all three cases so, keeping d constant, we observe that $E \propto V$ so larger the value of V larger the value of E will be so, calculated potential differences are related as ${V_1} > {V_3} > {V_2}$ hence, the order of electric field strength will also be in same order from highest to lowest.
Hence, order of electric field highest to lowest is ${E_1} > {E_3} > {E_2}.$

Note:
It should be remembered that the distance between the plates in each of the three cases remains the same which gives the derived order of strength of electric field and from the relation $V = Ed$ we can also say that electric field is the rate of change of potential difference with distance.