in spherical coordinates, using this coordinate system

we arrange the axes so that cup in question is symmetrically placed about origin O with the centre of the flat surface at O, and we place a unit test charge #Q_t# at O to calculate the field there.

The surface element for constant radius a follows as

#dS =a^{2} \sin theta \ d \theta \ d \varphi #

Assuming a uniform charge density #sigma # on the cup, the charge of a small element is

#dq = sigma dS# where #sigma = Q/A = Q/(2 pi a^2) # ,

the force due to a given small element will follow Couloumb's Law: #F = k (Q Q_t) / r^2#

#dF = k (dq Q_t) / a^2#

the force on the unit test charge #Q_t# placed at O will be zero due to symmetry. we need only look at the force in the z-direction. so here we can say that

#dF_z = k (dq Q_t) / a^2 * cos theta#

Putting it altogether

#dF_z = k ((sigma dS) Q_t) / a^2 * cos theta #

#(dF_z)/Q_t = dE_z = k sigma / a^2 \ cos theta \ dS#

#\implies dE_z = k sigma / a^2 \ cos theta \ a^{2} \sin theta \ d \theta \ d \varphi #

#= k sigma/2 \sin 2 theta \ d \theta \ d \varphi #

and as #k = 1/(4 pi epsilon_o) #

#\implies dE_z = 1/(4 pi epsilon_o) Q/(4 pi a^2) \sin 2 theta \ d \theta \ d \varphi #

# \implies dE_z = 1/((4 pi a)^2 epsilon_o) Q \sin 2 theta \ d \theta \ d \varphi #

# \implies E_z = 1/((4 pi a)^2 epsilon_o) Q int_{varphi = 0}^{2 pi} \ int_{theta = 0}^{pi/2} \ \sin 2 theta \ d \theta \ d \varphi #

# = 2 pi 1/((4 pi a)^2 epsilon_o) Q \ int_{theta = 0}^{pi/2} \ \sin 2 theta \ d \theta #

# = 1/( 8 pi epsilon_o a^2) Q [ -1/2 \cos 2 theta ]_{theta = 0}^{pi/2} \ #

# = 1/(8 pi epsilon_o a^2) Q [ \cos 2 theta ]_{theta = pi/2}^{0} \ #

# = Q/(8 pi epsilon_o a^2) #

# = (k Q)/(2 a^2 ) #