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Syllabus

C1=20 microfarad

C2= 40 microfarad

C3= 50 microfarad

$\mathbf{4}\mathbf{.}\mathbf{}Thechargeoncapacitor{C}_{1},\phantom{\rule{0ex}{0ex}}\left(A\right)0\mu C\left(B\right)5\mu C\left(C\right)10\mu C\left(D\right)Noneofthese\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{5}\mathbf{.}\mathbf{}Thechargeoncapacitor{C}_{2}\phantom{\rule{0ex}{0ex}}\left(A\right)0\mu C\left(B\right)5\mu C\left(C\right)10\mu C\left(D\right)Noneofthese\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{6}\mathbf{.}\mathbf{}Theworkdonebythebattery\phantom{\rule{0ex}{0ex}}\left(A\right)50\mu J\left(B\right)100\mu J\left(C\right)150\mu J\left(D\right)Zero$

1) V inner> V outer

2) V outer> V inner

3) V inner= V outer

4) both potential becomes zero

x-yplane is measured to have a magnitude 1.0 × 10^{–3}V/ m. What will be the magnitude of the electric field at the point(20 cm, 20 cm)(a) 5.0 × 10

^{–4 }V/m(b) 2.5 × 10

^{–4 }V/m(c) It will depend on the orientation of the dipole

(d) 1.25 × 10

^{–4 }V/mQ). Two capacitors of 3pF and 6pF are connected in series and a potential difference of 5000 V is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is

(a) 2250 V

(b) 2222 V

(c) 2.25$\times {10}^{6}$ V

(d) 1.1$\times {10}^{6}$ V

A cube of marble of each side 5m is placed in an electric field of intensity 300V/m . Determine the energy stored in marble , if its dielectric constant is 8.

$\left(a\right)\frac{1}{r}\phantom{\rule{0ex}{0ex}}\left(b\right)\frac{1}{{r}^{2}}\phantom{\rule{0ex}{0ex}}\left(c\right)\frac{1}{{r}^{3}}\phantom{\rule{0ex}{0ex}}\left(d\right){r}^{3}$

TNEI.(a) 28 / 9

(b) 4

(c) 5

(d) 18

a) 15 uF

b) 35 uF

c) 87.5 uF

d) 7.25 uF

It states that, "Total normal electric induction over a closed surface is equal to algebraic sum of total charges enclosed by the surface."

Pls explain it.

Common potential will be

^{-8}C hangs from a thread, making an angle of 30^{o}with the vertical, near a large conducting plate. Determine the surface charge density of the plate.In ans. it is gven that :

W=mg=(0.5*10

^{-6})*9.8 NBut m =10

^{-3}& M=10^{-6}Then how in the ans. it is given 10

^{-6}please explain as soon as possible.

A) 3 j B) 3.6j c) 5 j d)7.2 j

Q is zero then Q/q equals

Is Electrostatics meritnation study material equal to the study of maharashtraboard syllabus i.e., textbook? Plz answer fast as i m very confused between this two.....what is mean by hooks law ?

main ek essay

15 augustmain bolna chahta hu . toaap mughe lagbhag 600 words ka essay de sakte hai kya?microC CHARGE IS placed in it. An outward flux of O Is also given. it is req. that 6O FLUX ENTERS into the gaussian surface. calculate the amount of additional charge needed.in ans. it is given that:

According to the Question.:

2+Q= -60microCEPSILON 0

HOW? WHAT IS THE FORMULA?

please explain it as soon as possible.

Two long, thin, straight, parallel metal wires are separated by a distance 1m I air . The linear charge density on A is 5 * 10 ^ - 6 C/m , while that on B is -5*10 ^ - 6 C/m . Find the value of E at a point

(a) midway between them and (b) at a distance 0.5 m from A and 1.5m from B .

PLZ .. I NEED THIS URGENT

In Gauss's Law, why do we use and infinitely long cylinder carrying a charge +q ? What difference does it make by using an infinite conductor? Why not a finite one?